(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 60654, 810]*) (*NotebookOutlinePosition[ 61738, 847]*) (* CellTagsIndexPosition[ 61694, 843]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{Cell[TextData[ "showHeuristics\t=\tFalse;\nshowproof = True;\n\ntriangcount =\t0;\n\ triangtrue =\t0;\n\n(* 1.1 *)\ntrineq[3*csum[a*b]<=csum[a]^2];\n\ trineq[csum[a]^2<4*csum[a*b]];\n(* 1.2 *)\n\ trineq[csum[a^2]>=(36/35)*(s^2+a*b*c/s)];\n(* 1.3 *)\n\ trineq[8*cprod[s-a]<=a*b*c];\n(* 1.4 *)\ntrineq[8*a*b*c<=cprod[a+b]];\n(* 1.5 \ *)\ntrineq[3*cprod[a+b]<=8*csum[a^3]];\n(* 1.6 *)\n\ trineq[2*csum[a]*csum[a^2]>=3*(csum[a^3]+3*a*b*c)];\n(* 1.7 *)\n\ trineq[a*b*c=48*cprod[s-a]];\n(* 1.9 *)\n\ trineq[csum[a^3*(s-a)]<=a*b*c*s];\n(* 1.10 *)\n\ trineq[csum[a^5*(s-a)]<=(1/2)*a*b*c*csum[a^3]];\n(* 1.11 *)\n(*\n\ trineq[csum[a^2*b*(a-b)]>=0];\n*)\n(* 1.12 *)\n\ trineq[64*s^3*cprod[s-a]<=27*a^2*b^2*c^2];\n(* 1.14 *)\n\ trineq[2*s/(a*b*c)<=csum[1/a^2]];\n(* 1.15 *)\ntrineq[csum[1/(s-a)]>=9/s];\n\ (* 1.16 *)\ntrineq[3/2<=csum[a/(b+c)]];\ntrineq[csum[a/(b+c)]<2];\n(* 1.17 *)\ \ntrineq[15/4<=csum[(s+a)/(b+c)]];\ntrineq[csum[(s+a)/(b+c)]<9/2];\n(* 1.19 \ *)\ntrineq[1/3<=csum[a^2]/csum[a]^2];\ntrineq[csum[a^2]/csum[a]^2<1/2];\n(* \ 1.22 *)\ntrineq[csum[a^2]*csum[b^3*c^3]<2*csum[a^5]*csum[a^3]];\n(* 1.23 *)\n\ trineq[csum[a]^3<=5*csum[a*b*(a+b)]-3*a*b*c];\n\nPrint[\" \"];\n\ Print[\"Summary:\"];\nPrint[\"Proved \",triangtrue,\" inequalities out of \",\ \n\ttriangcount,\".\"];"], "Input", InitializationCell->True, AspectRatioFixed->True], Cell[TextData[ " \n 2\nTo prove: (a + b + c) >= 3 (a b + a c + b c)\n\ Expanding and collecting terms of the same sign gives:\n 2 2 2\na + b \ + c >= a b + a c + b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n \ 2 2 2\n(x + y) (x + z) (y + z)\n-------- + -------- + \ -------- >= \n 4 4 4\n \n (x + y) (x + z) (x + y) (y + \ z) (x + z) (y + z)\n --------------- + --------------- + ---------------\n \ 4 4 4\nMultiply both sides by 4 to \ get:\n 2 2 2\n2 (x + x y + y + x z + y z + z ) \ >= \n \n 2 2 2\n x + 3 x y + y + 3 x z + 3 \ y z + z\nExpanding and collecting terms of the same sign gives:\n 2 2 2\ \nx + y + z >= x y + x z + y z\nExpressing in terms of symmetric \ polynomials gives:\n[2,0,0] >= [1,1,0]\nThis result follows from the \ majorization theorem.\n \n 2\nTo \ prove: 4 (a b + a c + b c) > (a + b + c)\nExpanding and collecting terms of \ the same sign gives:\n 2 2 2\n2 a b + 2 a c + 2 \ b c > a + b + c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n (x + y) \ (x + z) (x + y) (y + z) (x + z) (y + z)\n2 (--------------- + \ --------------- + ---------------) > \n 4 4 \ 4\n \n 2 2 2\n (x + y) (x + z) (y + \ z)\n -------- + -------- + --------\n 4 4 4\nMultiply \ both sides by 2 to get:\n 2 2 2\nx + 3 x y + y \ + 3 x z + 3 y z + z > \n \n 2 2 2\n x + x y + y \ + x z + y z + z\nExpanding and collecting terms of the same sign gives:\n2 x \ y + 2 x z + 2 y z > 0\nThis is true because a sum of positive terms must be\\\ \n \n positive.\n \n a b c 2\n \ 36 (----- + s )\n 2 2 2 s\nTo prove: a + \ b + c >= ---------------\n 35\nSince 35 s is \ positive,\nMultiply both sides by 35 s to get:\n 2 2 2 \ 3\n35 (a + b + c ) s >= 36 (a b c + s )\nLet s=(a + b + c)/2.\nWe \ get:\n 2 2 2\n(a + b + c) (35 a + 35 b + 35 c )\ \n----------------------------------- >= \n 2\n \n \ 3\n 9 (a + b + c)\n 36 a b c + --------------\n \ 2\nMultiply both sides by 2 to get:\n 2 2 \ 2\n35 (a + b + c) (a + b + c ) >= \n \n 3 2 2 3 \ 2\n 9 (a + 3 a b + 3 a b + b + 3 a c + 14 a b c + \n \n 2 \ 2 2 3\n 3 b c + 3 a c + 3 b c + c )\nExpanding and \ collecting terms of the same sign gives:\n 3 2 2 3 \ 2 2\n26 a + 8 a b + 8 a b + 26 b + 8 a c + 8 b c + \n \n \ 2 2 3\n 8 a c + 8 b c + 26 c >= 126 a b c\nwe can divide \ both sides by 2 to get:\n 3 2 2 3 2 2\n13 \ a + 4 a b + 4 a b + 13 b + 4 a c + 4 b c + \n \n 2 2 \ 3\n 4 a c + 4 b c + 13 c >= 63 a b c\nLet a=(y+z)/2, b=(z+x)/2, \ c=(x+y)/2, to get:\n 3 2 2\n13 (x \ + y) (x + y) (x + z) (x + y) (x + z)\n----------- + ---------------- + \ ---------------- + \n 8 2 2\n \n \ 3 2 2\n 13 (x + z) (x + y) (y + z) (x + \ z) (y + z)\n ----------- + ---------------- + ---------------- + \n \ 8 2 2\n \n 2 \ 2 3\n (x + y) (y + z) (x + z) (y + z) 13 (y + z)\n \ ---------------- + ---------------- + ----------- >= \n 2 \ 2 8\n \n 63 (x + y) (x + z) (y + z)\n \ --------------------------\n 8\nMultiply both sides by 8 to get:\ \n 3 2 2 3 2\n34 x + 59 x y + 59 x y + 34 y \ + 59 x z + 48 x y z + \n \n 2 2 2 3\n 59 y \ z + 59 x z + 59 y z + 34 z >= \n \n 63 (x + y) (x + z) (y + z)\nExpanding \ and collecting terms of the same sign gives:\n 3 3 3\n34 x + \ 34 y + 34 z >= \n \n 2 2 2 2 2\ \n 4 x y + 4 x y + 4 x z + 78 x y z + 4 y z + 4 x z + \n \n 2\n \ 4 y z\nDividing both sides by 2 gives:\n 3 3 3\n17 x + 17 y \ + 17 z >= \n \n 2 2 2 2 2\n 2 \ x y + 2 x y + 2 x z + 39 x y z + 2 y z + 2 x z + \n \n 2\n 2 y \ z\nExpressing in terms of symmetric polynomials gives:\n17 [3,0,0] >= 13 \ [1,1,1] + 4 [2,1,0]\nThis follows from the following majorizations:\n13 \ [3,0,0] >= 13 [1,1,1]\n4 [3,0,0] >= 4 [2,1,0]\n \nTo prove: a b c >= 8 (-a + \ s) (-b + s) (-c + s)\nExpanding and collecting terms of the same sign gives:\n\ 2 2 2\n9 a b c + 8 a s + 8 b s + 8 c s >= \n \ \n 3\n 8 a b s + 8 a c s + 8 b c s + 8 s\n\ Let s=(a + b + c)/2.\nWe get:\n 2\n (a + b + c) \ (8 a + 8 b + 8 c)\n9 a b c + ------------------------------ >= \n \ 4\n \n 3 (a + b + c) (8 a b + 8 a c + 8 b c)\n (a + \ b + c) + -----------------------------------\n \ 2\nExpanding and collecting terms of the same sign gives:\n 3 3 \ 3\na + b + 3 a b c + c >= \n \n 2 2 2 2 2 \ 2\n a b + a b + a c + b c + a c + b c\nLet a=(y+z)/2, b=(z+x)/2, \ c=(x+y)/2, to get:\n 3 3\n(x + y) (x + z) 3 (x + y) (x + \ z) (y + z)\n-------- + -------- + ------------------------- + \n 8 \ 8 8\n \n 3 2 2\n\ (y + z) (x + y) (x + z) (x + y) (x + z)\n -------- >= \ ---------------- + ---------------- + \n 8 8 \ 8\n \n 2 2 2\n (x + \ y) (y + z) (x + z) (y + z) (x + y) (y + z)\n ---------------- + \ ---------------- + ---------------- + \n 8 8 \ 8\n \n 2\n (x + z) (y + z)\n ----------------\n\ 8\nMultiply both sides by 8 to get:\n 3 2 2 3 \ 2 2\n2 (x + 3 x y + 3 x y + y + 3 x z + 3 x y z + 3 \ y z + \n \n 2 2 3\n 3 x z + 3 y z + z ) >= \n \n \ 3 2 2 3 2\n 2 x + 5 x y + 5 x y + 2 y + 5 x z \ + 12 x y z + \n \n 2 2 2 3\n 5 y z + 5 x z + 5 \ y z + 2 z\nExpanding and collecting terms of the same sign gives:\n 2 \ 2 2 2 2 2\nx y + x y + x z + y z + x z + y z >= 6 x \ y z\nExpressing in terms of symmetric polynomials gives:\n[2,1,0] >= [1,1,1]\n\ This result follows from the majorization theorem.\n \nTo prove: (a + b) (a + \ c) (b + c) >= 8 a b c\nExpanding and collecting terms of the same sign gives:\ \n 2 2 2 2 2 2\na b + a b + a c + b c + a c + \ b c >= 6 a b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 2 \ 2 2\n(x + y) (x + z) (x + y) (x + z) (x + \ y) (y + z)\n---------------- + ---------------- + ---------------- + \n \ 8 8 8\n \n 2 \ 2 2\n (x + z) (y + z) (x + y) (y + z) (x + z) (y \ + z)\n ---------------- + ---------------- + ---------------- >\\\n \ 8 8 8\n \n 3 (x + y) (x + z) (y + z)\n \ = -------------------------\n 4\nMultiply both sides by 8 \ to get:\n 3 2 2 3 2\n2 x + 5 x y + 5 x y + 2 y \ + 5 x z + 12 x y z + \n \n 2 2 2 3\n 5 y z + 5 \ x z + 5 y z + 2 z >= \n \n 6 (x + y) (x + z) (y + z)\nExpanding and \ collecting terms of the same sign gives:\n 3 3 3 2 2 \ 2 2 2\n2 x + 2 y + 2 z >= x y + x y + x z + y z + x z + \ \n \n 2\n y z\nExpressing in terms of symmetric polynomials gives:\n\ [3,0,0] >= [2,1,0]\nThis result follows from the majorization theorem.\n \n \ 3 3 3\nTo prove: 8 (a + b + c ) >= 3 (a + b) (a + c) (b + \ c)\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 3 3 \ 3\n (x + y) (x + z) (y + z)\n8 (-------- + -------- + --------) \ >= \n 8 8 8\n \n 2 \ 2\n (x + y) (x + z) (x + y) (x + z)\n 3 (---------------- + \ ---------------- + \n 8 8\n \n 2\n \ (x + y) (y + z) (x + y) (x + z) (y + z)\n ---------------- + \ ----------------------- + \n 8 4\n \n \ 2 2 2\n (x + z) (y + z) \ (x + y) (y + z) (x + z) (y + z)\n ---------------- + ---------------- \ + ----------------)\n 8 8 8\n\ Multiply both sides by 8 to get:\n 3 2 2 3 2 \ 2\n8 (2 x + 3 x y + 3 x y + 2 y + 3 x z + 3 y z + \n \n 2 \ 2 3\n 3 x z + 3 y z + 2 z ) >= \n \n 3 2 \ 2 3 2\n 3 (2 x + 7 x y + 7 x y + 2 y + 7 x z + 16 x y z + \n \ \n 2 2 2 3\n 7 y z + 7 x z + 7 y z + 2 z )\ \nExpanding and collecting terms of the same sign gives:\n 3 2 \ 2 3 2 2\n10 x + 3 x y + 3 x y + 10 y + 3 x z + 3 y \ z + \n \n 2 2 3\n 3 x z + 3 y z + 10 z >= 48 x y z\n\ Expressing in terms of symmetric polynomials gives:\n3 [2,1,0] + 5 [3,0,0] >= \ 8 [1,1,1]\nThis follows from the following majorizations:\n5 [3,0,0] >= 5 \ [1,1,1]\n3 [2,1,0] >= 3 [1,1,1]\n \n 2 2 2\nTo \ prove: 2 (a + b + c) (a + b + c ) >= \n \n 3 3 3\n \ 3 (a + b + 3 a b c + c )\nExpanding and collecting terms of the same sign \ gives:\n 2 2 2 2 2 2\n2 a b + 2 a b \ + 2 a c + 2 b c + 2 a c + 2 b c >= \n \n 3 3 3\n a + \ b + 9 a b c + c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 2 \ 2 2\n (x + y) (x + z) (x + y) (x + z) \ (x + y) (y + z)\n2 (---------------- + ---------------- + ---------------- + \ \n 8 8 8\n \n 2 \ 2 2\n (x + z) (y + z) (x + y) (y + \ z) (x + z) (y + z)\n ---------------- + ---------------- + \ ----------------)\n 8 8 8\n \n \ 3 3\n (x + y) (x + z) 9 (x + y) (x + z) (y + \ z)\n >= -------- + -------- + ------------------------- + \n 8 \ 8 8\n \n 3\n (y + z)\n --------\n 8\ \nMultiply both sides by 4 to get:\n 3 2 2 3 2\n2 x \ + 5 x y + 5 x y + 2 y + 5 x z + 12 x y z + \n \n 2 2 \ 2 3\n 5 y z + 5 x z + 5 y z + 2 z >= \n \n 3 2 2 \ 3 2 2\n x + 6 x y + 6 x y + y + 6 x z + 9 x y z \ + 6 y z + \n \n 2 2 3\n 6 x z + 6 y z + z\nExpanding \ and collecting terms of the same sign gives:\n 3 3 3\nx + y \ + 3 x y z + z >= \n \n 2 2 2 2 2 2\n x y + x \ y + x z + y z + x z + y z\nExpressing in terms of symmetric polynomials \ gives:\n[1,1,1] + [3,0,0] >= 2 [2,1,0]\nThis follows from the following \ majorizations:\n[1,1,1] + [3,0,0] >= 2 [2,1,0]\n \n 2 2 \ 2\nTo prove: a (-a + s) + b (-b + s) + c (-c + s) > a b c\n\ Expanding and collecting terms of the same sign gives:\n 2 2 2 \ 3 3 3\na s + b s + c s > a + b + a b c + c\nLet s=(a + b + \ c)/2.\nWe get:\n 2 2 2\n(a + b + c) (a + b + c ) 3 \ 3 3\n-------------------------- > a + b + a b c + c\n \ 2\nMultiply both sides by 2 to get:\n 2 2 2 3 3 \ 3\n(a + b + c) (a + b + c ) > 2 (a + b + a b c + c )\nExpanding \ and collecting terms of the same sign gives:\n 2 2 2 2 \ 2 2\na b + a b + a c + b c + a c + b c > \n \n 3 3 \ 3\n a + b + 2 a b c + c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n \ 2 2 2\n(x + y) (x + z) (x + y) (x + \ z) (x + y) (y + z)\n---------------- + ---------------- + \ ---------------- + \n 8 8 8\n \n \ 2 2 2\n (x + z) (y + z) (x \ + y) (y + z) (x + z) (y + z)\n ---------------- + ---------------- + \ ---------------- > \n 8 8 8\n \n \ 3 3 3\n (x + y) (x + z) \ (x + y) (x + z) (y + z) (y + z)\n -------- + -------- + \ ----------------------- + --------\n 8 8 4 \ 8\nMultiply both sides by 8 to get:\n 3 2 2 3 \ 2\n2 x + 5 x y + 5 x y + 2 y + 5 x z + 12 x y z + \n \n 2 \ 2 2 3\n 5 y z + 5 x z + 5 y z + 2 z > \n \n 3 \ 2 2 3 2\n 2 x + 5 x y + 5 x y + 2 y + 5 x z + 4 x y \ z + \n \n 2 2 2 3\n 5 y z + 5 x z + 5 y z + 2 \ z\nExpanding and collecting terms of the same sign gives:\n8 x y z > 0\nThis \ is true because a sum of positive terms must be\\\n \n positive.\n \n \ 3 a b c 2 2 2\nTo prove: ------- >= a (-a + s) \ + b (-b + s) + c (-c + s)\n 2\nMultiply both sides by 2 to get:\ \n 3 3 3 2 2 2\n3 a b c >= 2 (-a - b - c \ + a s + b s + c s)\nExpanding and collecting terms of the same sign gives:\ \n 3 3 3 2 2 2\n2 a + 2 b + 3 a b \ c + 2 c >= 2 a s + 2 b s + 2 c s\nLet s=(a + b + c)/2.\nWe get:\n 3 \ 3 3\n2 a + 2 b + 3 a b c + 2 c >= \n \n 2 \ 2 2\n (a + b + c) (2 a + 2 b + 2 c )\n \ --------------------------------\n 2\nExpanding and \ collecting terms of the same sign gives:\n 3 3 3\na + b + 3 \ a b c + c >= \n \n 2 2 2 2 2 2\n a b + a b + \ a c + b c + a c + b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n \ 3 3\n(x + y) (x + z) 3 (x + y) (x + z) (y + z)\n-------- + \ -------- + ------------------------- + \n 8 8 8\n\ \n 3 2 2\n (y + z) (x + y) \ (x + z) (x + y) (x + z)\n -------- >= ---------------- + \ ---------------- + \n 8 8 8\n \n \ 2 2 2\n (x + y) (y + z) (x + \ z) (y + z) (x + y) (y + z)\n ---------------- + ---------------- + \ ---------------- + \n 8 8 8\n \n \ 2\n (x + z) (y + z)\n ----------------\n 8\n\ Multiply both sides by 8 to get:\n 3 2 2 3 2 \ 2\n2 (x + 3 x y + 3 x y + y + 3 x z + 3 x y z + 3 y z + \n \n \ 2 2 3\n 3 x z + 3 y z + z ) >= \n \n 3 2 \ 2 3 2\n 2 x + 5 x y + 5 x y + 2 y + 5 x z + 12 x y z + \n\ \n 2 2 2 3\n 5 y z + 5 x z + 5 y z + 2 z\n\ Expanding and collecting terms of the same sign gives:\n 2 2 2 \ 2 2 2\nx y + x y + x z + y z + x z + y z >= 6 x y z\n\ Expressing in terms of symmetric polynomials gives:\n[2,1,0] >= [1,1,1]\nThis \ result follows from the majorization theorem.\n \nTo prove: a b (a + b) + a c \ (a + c) + b c (b + c) >= \n \n 48 (-a + s) (-b + s) (-c + s)\nExpanding and \ collecting terms of the same sign gives:\n 2 2 2 2 \ 2 2\na b + a b + a c + 48 a b c + b c + a c + b c + \n \n \ 2 2 2\n 48 a s + 48 b s + 48 c s >= \n \n \ 3\n 48 a b s + 48 a c s + 48 b c s + 48 s\nLet \ s=(a + b + c)/2.\nWe get:\n 2 2 2 2 2 2\ \na b + a b + a c + 48 a b c + b c + a c + b c + \n \n 2\n \ (a + b + c) (48 a + 48 b + 48 c)\n --------------------------------- >= \ \n 4\n \n 3 (a + b + c) (48 a b + 48 a c + \ 48 b c)\n 6 (a + b + c) + --------------------------------------\n \ 2\nExpanding and collecting terms of the same sign \ gives:\n 3 3 3\n6 a + 6 b + 12 a b c + 6 c >= \n \n \ 2 2 2 2 2 2\n 5 a b + 5 a b + 5 a \ c + 5 b c + 5 a c + 5 b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n \ 3 3\n (x + y) (x + z) (x + y) (x + z) (y + z)\n6 \ (-------- + -------- + ----------------------- + \n 8 8 \ 4\n \n 3 2 2\n \ (y + z) (x + y) (x + z) (x + y) (x + z)\n --------) >= 5 \ (---------------- + ---------------- + \n 8 8 \ 8\n \n 2 2\n (x + y) (y + z) (x \ + z) (y + z)\n ---------------- + ---------------- + \n 8 \ 8\n \n 2 2\n (x + y) (y \ + z) (x + z) (y + z)\n ---------------- + ----------------)\n \ 8 8\nMultiply both sides by 8 to get:\n 3 2 \ 2 3 2\n6 (2 x + 5 x y + 5 x y + 2 y + 5 x z + 4 x y z + \n \ \n 2 2 2 3\n 5 y z + 5 x z + 5 y z + 2 z ) \ >= \n \n 3 2 2 3 2\n 5 (2 x + 5 x y + 5 x y \ + 2 y + 5 x z + 12 x y z + \n \n 2 2 2 3\n \ 5 y z + 5 x z + 5 y z + 2 z )\nExpanding and collecting terms of the same \ sign gives:\n 3 2 2 3 2 2 2\n2 x + \ 5 x y + 5 x y + 2 y + 5 x z + 5 y z + 5 x z + \n \n 2 3\n \ 5 y z + 2 z >= 36 x y z\nExpressing in terms of symmetric polynomials \ gives:\n5 [2,1,0] + [3,0,0] >= 6 [1,1,1]\nThis follows from the following \ majorizations:\n[3,0,0] >= [1,1,1]\n5 [2,1,0] >= 5 [1,1,1]\n \n \ 3 3 3\nTo prove: a b c s >= a (-a + s) + b \ (-b + s) + c (-c + s)\nExpanding and collecting terms of the same sign \ gives:\n 4 4 4 3 3 3\na + b + c + a b c s >= \ a s + b s + c s\nLet s=(a + b + c)/2.\nWe get:\n 4 4 4 a b c (a + \ b + c)\na + b + c + ----------------- >= \n 2\n \n \ 3 3 3\n (a + b + c) (a + b + c )\n \ --------------------------\n 2\nMultiply both sides by 2 to get:\ \n 4 4 2 2 2 4\n2 a + 2 b + a b c + a b \ c + a b c + 2 c >= \n \n 3 3 3\n (a + b + c) (a + b \ + c )\nExpanding and collecting terms of the same sign gives:\n 4 4 2 \ 2 2 4\na + b + a b c + a b c + a b c + c >= \n \n \ 3 3 3 3 3 3\n a b + a b + a c + b c + a c + \ b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 4 4 \ 2\n(x + y) (x + z) (x + y) (x + z) (y + z)\n-------- + -------- + \ ------------------------ + \n 16 16 16\n \n \ 2 2\n (x + y) (x + z) (y + z) \ (x + y) (x + z) (y + z)\n ------------------------ + \ ------------------------ + \n 16 16\n \n \ 4 3 3\n (y + z) (x + y) (x \ + z) (x + y) (x + z)\n -------- >= ---------------- + ---------------- + \ \n 16 16 16\n \n 3 \ 3 3\n (x + y) (y + z) (x + z) (y + z) (x + \ y) (y + z)\n ---------------- + ---------------- + ---------------- + \n \ 16 16 16\n \n 3\n (x \ + z) (y + z)\n ----------------\n 16\nMultiply both sides by 8 to \ get:\n 4 3 2 2 3 4 3 2\nx + 3 x y + 5 x \ y + 3 x y + y + 3 x z + 4 x y z + \n \n 2 3 2 2 \ 2 2 2\n 4 x y z + 3 y z + 5 x z + 4 x y z + 5 y z + \n \n\ 3 3 4\n 3 x z + 3 y z + z >= \n \n 4 3 2 \ 2 3 4 3 2\n x + 3 x y + 3 x y + 3 x y + y + 3 x \ z + 6 x y z + \n \n 2 3 2 2 2 2 2\n \ 6 x y z + 3 y z + 3 x z + 6 x y z + 3 y z + \n \n 3 3 \ 4\n 3 x z + 3 y z + z\nExpanding and collecting terms of the same sign \ gives:\n 2 2 2 2 2 2\n2 x y + 2 x z + 2 y z >= \n \n \ 2 2 2\n 2 x y z + 2 x y z + 2 x y z\nDividing both \ sides by 2 gives:\n 2 2 2 2 2 2 2 2 2\nx y + \ x z + y z >= x y z + x y z + x y z\nExpressing in terms of symmetric \ polynomials gives:\n[2,2,0] >= [2,1,1]\nThis result follows from the \ majorization theorem.\n \n 3 3 3\n a b c (a \ + b + c )\nTo prove: -------------------- >= \n 2\n \n \ 5 5 5\n a (-a + s) + b (-b + s) + c (-c + s)\n\ Multiply both sides by 2 to get:\n 3 3 3\na b c (a + b + c ) \ >= 2 \n \n 6 6 6 5 5 5\n (-a - b - c + a s + b \ s + c s)\nExpanding and collecting terms of the same sign gives:\n 6 \ 6 4 4 4 6\n2 a + 2 b + a b c + a b c + a b c + \ 2 c >= \n \n 5 5 5\n 2 a s + 2 b s + 2 c s\nLet s=(a + \ b + c)/2.\nWe get:\n 6 6 4 4 4 6\n2 a + 2 b \ + a b c + a b c + a b c + 2 c >= \n \n 5 5 5\n\ (a + b + c) (2 a + 2 b + 2 c )\n --------------------------------\n \ 2\nExpanding and collecting terms of the same sign gives:\n 6 6 \ 4 4 4 6\na + b + a b c + a b c + a b c + c >= \n\ \n 5 5 5 5 5 5\n a b + a b + a c + b c + a \ c + b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 6 6 \ 4\n(x + y) (x + z) (x + y) (x + z) (y + z)\n-------- + -------- \ + ------------------------ + \n 64 64 64\n \n \ 4 4\n (x + y) (x + z) (y + z) \ (x + y) (x + z) (y + z)\n ------------------------ + \ ------------------------ + \n 64 64\n \n \ 6 5 5\n (y + z) (x + y) (x \ + z) (x + y) (x + z)\n -------- >= ---------------- + ---------------- + \ \n 64 64 64\n \n 5 \ 5 5\n (x + y) (y + z) (x + z) (y + z) (x + \ y) (y + z)\n ---------------- + ---------------- + ---------------- + \n \ 64 64 64\n \n 5\n (x \ + z) (y + z)\n ----------------\n 64\nMultiply both sides by 64 to \ get:\n 6 5 4 2 3 3 2 4 5\n2 (x + 4 x \ y + 10 x y + 13 x y + 10 x y + 4 x y + \n \n 6 5 4 \ 3 2 2 3\n y + 4 x z + 5 x y z + 7 x y z + 7 x y z + \ \n \n 4 5 4 2 3 2\n 5 x y z + 4 y z + \ 10 x z + 7 x y z + \n \n 2 2 2 3 2 4 2 3 3\ \n 9 x y z + 7 x y z + 10 y z + 13 x z + \n \n 2 3 \ 2 3 3 3 2 4\n 7 x y z + 7 x y z + 13 y z + 10 x \ z + \n \n 4 2 4 5 5 6\n 5 x y z + 10 \ y z + 4 x z + 4 y z + z ) >= \n \n 6 5 4 2 3 3 \ 2 4 5\n 2 x + 8 x y + 15 x y + 20 x y + 15 x y + 8 x y \ + \n \n 6 5 4 3 2 2 3\n 2 y + 8 x z \ + 20 x y z + 20 x y z + 20 x y z + \n \n 4 5 4 2 \ 3 2\n 20 x y z + 8 y z + 15 x z + 20 x y z + \n \n 3 \ 2 4 2 3 3 2 3\n 20 x y z + 15 y z + 20 x z + \ 20 x y z + \n \n 2 3 3 3 2 4 4\n 20 x y \ z + 20 y z + 15 x z + 20 x y z + \n \n 2 4 5 5 \ 6\n 15 y z + 8 x z + 8 y z + 2 z\nExpanding and collecting terms of \ the same sign gives:\n 4 2 3 3 2 4 4 2 2 2 2\n5 \ x y + 6 x y + 5 x y + 5 x z + 18 x y z + \n \n 4 2 3 3 \ 3 3 2 4 2 4\n 5 y z + 6 x z + 6 y z + 5 x z + 5 y \ z >= \n \n 4 3 2 2 3 4\n 10 x y z + 6 x \ y z + 6 x y z + 10 x y z + \n \n 3 2 3 2 2 3 \ 2 3 4\n 6 x y z + 6 x y z + 6 x y z + 6 x y z + 10 x y \ z\nExpressing in terms of symmetric polynomials gives:\n3 [2,2,2] + 3 [3,3,0] \ + 5 [4,2,0] >= 6 [3,2,1] + 5 [4,1,1]\nThis follows from the following \ majorizations:\n3 [2,2,2] + 3 [3,3,0] >= 6 [3,2,1]\n5 [4,2,0] >= 5 [4,1,1]\n \ \n 2 2 2 3\nTo prove: 27 a b c >= 64 s (-a + s) (-b \ + s) (-c + s)\nExpanding and collecting terms of the same sign gives:\n 2 \ 2 2 3 5 5 5\n27 a b c + 64 a b c s + \ 64 a s + 64 b s + 64 c s >= \n \n 4 4 4 \ 6\n 64 a b s + 64 a c s + 64 b c s + 64 s\nLet s=(a + b + c)/2.\nWe get:\n\ 2 2 2 3\n27 a b c + 8 a b c (a + b + c) + \n \n\ 5\n (a + b + c) (64 a + 64 b + 64 c)\n \ --------------------------------- >= \n 32\n \n \ 4\n 6 (a + b + c) (64 a b + 64 a c + 64 b c)\n \ (a + b + c) + ---------------------------------------\n \ 16\nExpanding and collecting terms of the same sign gives:\n 6 \ 5 5 6 5 4 4\na + 2 a b + 2 a b + b \ + 2 a c + 2 a b c + 2 a b c + \n \n 5 2 2 2 4 \ 5 5 6\n 2 b c + 21 a b c + 2 a b c + 2 a c + 2 b c + c \ >\\\n \n 4 2 3 3 2 4 3 2 2 3\n = a b + 4 a \ b + a b + 4 a b c + 4 a b c + \n \n 4 2 3 2 3 2 \ 4 2 3 3\n a c + 4 a b c + 4 a b c + b c + 4 a c + \n \n \ 2 3 2 3 3 3 2 4 2 4\n 4 a b c + 4 a b c + 4 \ b c + a c + b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 6 \ 5 5 6\n(x + y) (x + y) (x + z) \ (x + y) (x + z) (x + z)\n-------- + ---------------- + ---------------- + \ -------- + \n 64 32 32 64\n \n \ 5 4\n (x + y) (y + z) (x + y) (x + z) (y + z)\n \ ---------------- + ------------------------ + \n 32 \ 32\n \n 4 5\n (x + y) (x + z) (y + z) \ (x + z) (y + z)\n ------------------------ + ---------------- + \n \ 32 32\n \n 2 2 2\n 21 \ (x + y) (x + z) (y + z)\n ----------------------------- + \n \ 64\n \n 4 5\n (x + y) (x + z) \ (y + z) (x + y) (y + z)\n ------------------------ + ---------------- + \ \n 32 32\n \n 5 6\n\ (x + z) (y + z) (y + z)\n ---------------- + -------- >= \n \ 32 64\n \n 4 2 3 3\n (x + y) (x \ + z) (x + y) (x + z)\n ----------------- + ----------------- + \n \ 64 16\n \n 2 4 3 2\n (x + \ y) (x + z) (x + y) (x + z) (y + z)\n ----------------- + \ ------------------------- + \n 64 16\n \n \ 2 3 4 2\n (x + y) (x + z) (y + z) (x \ + y) (y + z)\n ------------------------- + ----------------- + \n \ 16 64\n \n 3 2 \ 3 2\n (x + y) (x + z) (y + z) (x + y) (x + z) (y + z)\n \ ------------------------- + ------------------------- + \n 16 \ 16\n \n 4 2 3 3\n (x \ + z) (y + z) (x + y) (y + z)\n ----------------- + ----------------- + \ \n 64 16\n \n 2 3 \ 2 3\n (x + y) (x + z) (y + z) (x + y) (x + z) (y + z)\n \ ------------------------- + ------------------------- + \n 16 \ 16\n \n 3 3 2 4 \ 2 4\n (x + z) (y + z) (x + y) (y + z) (x + z) (y + z)\n \ ----------------- + ----------------- + -----------------\n 16 \ 64 64\nMultiply both sides by 64 to get:\n 6 \ 5 4 2 3 3 2 4\n2 (3 x + 13 x y + 38 x y + 57 \ x y + 38 x y + \n \n 5 6 5 4 3 2\n\ 13 x y + 3 y + 13 x z + 51 x y z + 97 x y z + \n \n 2 3 \ 4 5 4 2\n 97 x y z + 51 x y z + 13 y z + 38 \ x z + \n \n 3 2 2 2 2 3 2 4 2\n 97 \ x y z + 123 x y z + 97 x y z + 38 y z + \n \n 3 3 2 \ 3 2 3 3 3\n 57 x z + 97 x y z + 97 x y z + 57 y z \ + \n \n 2 4 4 2 4 5 5\n 38 x \ z + 51 x y z + 38 y z + 13 x z + 13 y z + \n \n 6 6 \ 5 4 2 3 3\n 3 z ) >= 6 x + 26 x y + 49 x y + 60 x y \ + \n \n 2 4 5 6 5 4\n 49 x y + 26 x y \ + 6 y + 26 x z + 112 x y z + \n \n 3 2 2 3 4 \ 5\n 224 x y z + 224 x y z + 112 x y z + 26 y z + \n \n 4 \ 2 3 2 2 2 2 3 2\n 49 x z + 224 x y z + \ 360 x y z + 224 x y z + \n \n 4 2 3 3 2 3 \ 2 3\n 49 y z + 60 x z + 224 x y z + 224 x y z + \n \n 3 3 \ 2 4 4 2 4 5\n 60 y z + 49 x z + 112 x \ y z + 49 y z + 26 x z + \n \n 5 6\n 26 y z + 6 z\n\ Expanding and collecting terms of the same sign gives:\n 4 2 3 3 \ 2 4 4 2 4 2\n27 x y + 54 x y + 27 x y + 27 x z + \ 27 y z + \n \n 3 3 3 3 2 4 2 4\n 54 x z + \ 54 y z + 27 x z + 27 y z >= \n \n 4 3 2 2 3 \ 4\n 10 x y z + 30 x y z + 30 x y z + 10 x y z + \n \n 3 \ 2 2 2 2 3 2 2 3\n 30 x y z + 114 x y z + \ 30 x y z + 30 x y z + \n \n 2 3 4\n 30 x y z + 10 \ x y z\nExpressing in terms of symmetric polynomials gives:\n27 [3,3,0] + 27 \ [4,2,0] >= \n \n 19 [2,2,2] + 30 [3,2,1] + 5 [4,1,1]\nThis follows from the \ following majorizations:\n5 [4,2,0] >= 5 [4,1,1]\n27 [3,3,0] >= 27 [3,2,1]\n3 \ [4,2,0] >= 3 [3,2,1]\n19 [4,2,0] >= 19 [2,2,2]\n \n -2 -2 -2 \ 2 s\nTo prove: a + b + c >= -----\n a b \ c\n 2 2 2\nSince a b c is positive,\n 2 2 \ 2\nMultiply both sides by a b c to get:\n 2 2 2 2 2 2\na b + a \ c + b c >= 2 a b c s\nLet s=(a + b + c)/2.\nWe get:\n 2 2 2 2 2 2\ \na b + a c + b c >= a b c (a + b + c)\nLet a=(y+z)/2, b=(z+x)/2, \ c=(x+y)/2, to get:\n 2 2 2 2\n(x + y) (x + z) \ (x + y) (y + z)\n----------------- + ----------------- + \n 16 \ 16\n \n 2 2 2\n (x + z) (y + z) (x \ + y) (x + z) (y + z)\n ----------------- >= ------------------------ + \n \ 16 16\n \n 2 \ 2\n (x + y) (x + z) (y + z) (x + y) (x + z) (y + z)\n \ ------------------------ + ------------------------\n 16 \ 16\nMultiply both sides by 16 to get:\n 4 3 2 2 \ 3 4 3 2\nx + 2 x y + 3 x y + 2 x y + y + 2 x z + \ 8 x y z + \n \n 2 3 2 2 2 2 2\n 8 x y \ z + 2 y z + 3 x z + 8 x y z + 3 y z + \n \n 3 3 4\n \ 2 x z + 2 y z + z >= \n \n 3 2 2 3 3 2 \ 2\n 2 (x y + 2 x y + x y + x z + 4 x y z + 4 x y z + \n \n 3 \ 2 2 2 2 2 3 3\n y z + 2 x z + 4 x y z \ + 2 y z + x z + y z )\nExpanding and collecting terms of the same sign \ gives:\n 4 4 4 2 2 2 2 2 2\nx + y + z >= x y + x z \ + y z\nExpressing in terms of symmetric polynomials gives:\n[4,0,0] >= \ [2,2,0]\nThis result follows from the majorization theorem.\n \n 1 \ 1 1 9\nTo prove: ------ + ------ + ------ >= -\n \ -a + s -b + s -c + s s\nSince (a - s) s (-b + s) (-c + s) is negative,\ \nMultiply both sides by (b - s) (c - s) s (-a + s) to get:\n \ 2\ns (a b + a c + b c - 2 a s - 2 b s - 2 c s \ + 3 s ) >= \n \n 9 (a - s) (c - s) (-b + s)\nExpanding and collecting terms \ of the same sign gives:\n 2 2 2\n9 a b c + 7 a s \ + 7 b s + 7 c s >= \n \n 3\n 8 a b s + 8 \ a c s + 8 b c s + 6 s\nLet s=(a + b + c)/2.\nWe get:\n 2\n\ (a + b + c) (7 a + 7 b + 7 c)\n9 a b c + \ ------------------------------ >= \n 4\n \n \ 3\n 3 (a + b + c) (a + b + c) (8 a b + 8 a c + 8 b c)\n \ -------------- + -----------------------------------\n 4 \ 2\nMultiply both sides by 4 to get:\n 3 2 2 \ 3 2\n7 a + 21 a b + 21 a b + 7 b + 21 a c + 78 a b c + \n \n \ 2 2 2 3\n 21 b c + 21 a c + 21 b c + 7 c >= \n \ \n 3 2 2 3 2\n 3 a + 25 a b + 25 a b + 3 b \ + 25 a c + 66 a b c + \n \n 2 2 2 3\n 25 b \ c + 25 a c + 25 b c + 3 c\nExpanding and collecting terms of the same sign \ gives:\n 3 3 3\n4 a + 4 b + 12 a b c + 4 c >= \n \n \ 2 2 2 2 2 2\n 4 a b + 4 a b + 4 a \ c + 4 b c + 4 a c + 4 b c\nwe can divide both sides by 4 to get:\n 3 3 \ 3\na + b + 3 a b c + c >= \n \n 2 2 2 2 \ 2 2\n a b + a b + a c + b c + a c + b c\nLet a=(y+z)/2, \ b=(z+x)/2, c=(x+y)/2, to get:\n 3 3\n(x + y) (x + z) 3 \ (x + y) (x + z) (y + z)\n-------- + -------- + ------------------------- + \n \ 8 8 8\n \n 3 2 \ 2\n (y + z) (x + y) (x + z) (x + y) (x + z)\n -------- \ >= ---------------- + ---------------- + \n 8 8 \ 8\n \n 2 2 2\n (x \ + y) (y + z) (x + z) (y + z) (x + y) (y + z)\n ---------------- + \ ---------------- + ---------------- + \n 8 8 \ 8\n \n 2\n (x + z) (y + z)\n ----------------\n\ 8\nMultiply both sides by 8 to get:\n 3 2 2 3 \ 2 2\n2 (x + 3 x y + 3 x y + y + 3 x z + 3 x y z + 3 \ y z + \n \n 2 2 3\n 3 x z + 3 y z + z ) >= \n \n \ 3 2 2 3 2\n 2 x + 5 x y + 5 x y + 2 y + 5 x z \ + 12 x y z + \n \n 2 2 2 3\n 5 y z + 5 x z + 5 \ y z + 2 z\nExpanding and collecting terms of the same sign gives:\n 2 \ 2 2 2 2 2\nx y + x y + x z + y z + x z + y z >= 6 x \ y z\nExpressing in terms of symmetric polynomials gives:\n[2,1,0] >= [1,1,1]\n\ This result follows from the majorization theorem.\n \n c b \ a 3\nTo prove: ----- + ----- + ----- >= -\n a + b a + c \ b + c 2\nSince 2 (a + b) (a + c) (b + c) is positive,\nMultiply both \ sides by 2 (a + b) (a + c) (b + c) to get:\n 3 2 2 3 2 \ 2 2\n2 (a + a b + a b + b + a c + 3 a b c + b c + a c \ + \n \n 2 3\n b c + c ) >= 3 (a + b) (a + c) (b + c)\n\ Expanding and collecting terms of the same sign gives:\n 3 3 3 \ 2 2 2 2 2\n2 a + 2 b + 2 c >= a b + a b + a c + \ b c + a c + \n \n 2\n b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to \ get:\n 3 3 3\n (x + y) (x + z) (y + z)\n2 \ (-------- + -------- + --------) >= \n 8 8 8\n \n \ 2 2 2\n (x + y) (x + z) (x + y) (x + \ z) (x + y) (y + z)\n ---------------- + ---------------- + \ ---------------- + \n 8 8 8\n \n \ 2 2 2\n (x + z) (y + z) \ (x + y) (y + z) (x + z) (y + z)\n ---------------- + ---------------- + \ ----------------\n 8 8 8\nMultiply \ both sides by 8 to get:\n 3 2 2 3 2 2\n2 \ (2 x + 3 x y + 3 x y + 2 y + 3 x z + 3 y z + \n \n 2 2 \ 3\n 3 x z + 3 y z + 2 z ) >= \n \n 3 2 2 3 \ 2\n 2 x + 5 x y + 5 x y + 2 y + 5 x z + 12 x y z + \n \n 2 \ 2 2 3\n 5 y z + 5 x z + 5 y z + 2 z\nExpanding and \ collecting terms of the same sign gives:\n 3 2 2 3 2 \ 2 2 2\n2 x + x y + x y + 2 y + x z + y z + x z + y z + \n \ \n 3\n 2 z >= 12 x y z\nExpressing in terms of symmetric polynomials \ gives:\n[2,1,0] + [3,0,0] >= 2 [1,1,1]\nThis follows from the following \ majorizations:\n[3,0,0] >= [1,1,1]\n[2,1,0] >= [1,1,1]\n \n c \ b a\nTo prove: 2 > ----- + ----- + -----\n a + b a \ + c b + c\nSince (a + b) (a + c) (b + c) is positive,\nMultiply both sides \ by (a + b) (a + c) (b + c) to get:\n2 (a + b) (a + c) (b + c) > \n \n 3 \ 2 2 3 2 2 2\n a + a b + a b + b + a \ c + 3 a b c + b c + a c + \n \n 2 3\n b c + c\nExpanding and \ collecting terms of the same sign gives:\n 2 2 2 2 \ 2 2\na b + a b + a c + a b c + b c + a c + b c > \n \n 3 3 \ 3\n a + b + c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 2 \ 2 2\n(x + y) (x + z) (x + y) (x + z) (x \ + y) (y + z)\n---------------- + ---------------- + ---------------- + \n \ 8 8 8\n \n \ 2\n (x + y) (x + z) (y + z) (x + z) (y + z)\n \ ----------------------- + ---------------- + \n 8 \ 8\n \n 2 2\n (x + y) (y + z) (x + \ z) (y + z)\n ---------------- + ---------------- > \n 8 \ 8\n \n 3 3 3\n (x + y) (x + z) (y + z)\ \n -------- + -------- + --------\n 8 8 8\nMultiply \ both sides by 8 to get:\n 3 2 2 3 2 \ 2\n2 (x + 3 x y + 3 x y + y + 3 x z + 7 x y z + 3 y z + \n \n \ 2 2 3\n 3 x z + 3 y z + z ) > \n \n 3 2 2 \ 3 2 2\n 2 x + 3 x y + 3 x y + 2 y + 3 x z + 3 y z + \n \ \n 2 2 3\n 3 x z + 3 y z + 2 z\nExpanding and \ collecting terms of the same sign gives:\n 2 2 2 \ 2 2\n3 x y + 3 x y + 3 x z + 14 x y z + 3 y z + 3 x z + \n \ \n 2\n 3 y z > 0\nThis is true because a sum of positive terms must \ be\\\n \n positive.\n \n a + s b + s c + s 15\nTo prove: \ ----- + ----- + ----- >= --\n b + c a + c a + b 4\nSince 4 (a \ + b) (a + c) (b + c) is positive,\nMultiply both sides by 4 (a + b) (a + c) \ (b + c) to get:\n 3 2 2 3 2 2 2\n4 \ (a + a b + a b + b + a c + 3 a b c + b c + a c + \n \n 2 3 \ 2 2\n b c + c + a s + 3 a b s + b s + 3 a c s + \n \n \ 2\n 3 b c s + c s) >= 15 (a + b) (a + c) (b + c)\n\ Expanding and collecting terms of the same sign gives:\n 3 3 3 \ 2 2\n4 a + 4 b + 4 c + 4 a s + 12 a b s + 4 b s + \n \ \n 2\n 12 a c s + 12 b c s + 4 c s >= \n \n \ 2 2 2 2\n 11 a b + 11 a b + 11 a c + \ 18 a b c + 11 b c + \n \n 2 2\n 11 a c + 11 b c\nLet s=(a \ + b + c)/2.\nWe get:\n 3 3 3\n4 a + 4 b + 4 c + ((a + b + c) \n\ \n 2 2 2\n (4 a + 12 a b \ + 4 b + 12 a c + 12 b c + 4 c )) / 2 \\\n \n 2 2 2 \ 2\n >= 11 a b + 11 a b + 11 a c + 18 a b c + 11 b c + \ \n \n 2 2\n 11 a c + 11 b c\nExpanding and collecting \ terms of the same sign gives:\n 3 3 3 2 2 2 \ 2\n6 a + 6 b + 6 c >= 3 a b + 3 a b + 3 a c + 3 b c + \n \n \ 2 2\n 3 a c + 3 b c\nwe can divide both sides by 3 to get:\n 3 \ 3 3 2 2 2 2 2 2\n2 (a + b + c ) >= a \ b + a b + a c + b c + a c + b c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to \ get:\n 3 3 3\n (x + y) (x + z) (y + z)\n2 \ (-------- + -------- + --------) >= \n 8 8 8\n \n \ 2 2 2\n (x + y) (x + z) (x + y) (x + \ z) (x + y) (y + z)\n ---------------- + ---------------- + \ ---------------- + \n 8 8 8\n \n \ 2 2 2\n (x + z) (y + z) \ (x + y) (y + z) (x + z) (y + z)\n ---------------- + ---------------- + \ ----------------\n 8 8 8\nMultiply \ both sides by 8 to get:\n 3 2 2 3 2 2\n2 \ (2 x + 3 x y + 3 x y + 2 y + 3 x z + 3 y z + \n \n 2 2 \ 3\n 3 x z + 3 y z + 2 z ) >= \n \n 3 2 2 3 \ 2\n 2 x + 5 x y + 5 x y + 2 y + 5 x z + 12 x y z + \n \n 2 \ 2 2 3\n 5 y z + 5 x z + 5 y z + 2 z\nExpanding and \ collecting terms of the same sign gives:\n 3 2 2 3 2 \ 2 2 2\n2 x + x y + x y + 2 y + x z + y z + x z + y z + \n \ \n 3\n 2 z >= 12 x y z\nExpressing in terms of symmetric polynomials \ gives:\n[2,1,0] + [3,0,0] >= 2 [1,1,1]\nThis follows from the following \ majorizations:\n[3,0,0] >= [1,1,1]\n[2,1,0] >= [1,1,1]\n \n 9 a + \ s b + s c + s\nTo prove: - > ----- + ----- + -----\n 2 b + c \ a + c a + b\nSince 2 (a + b) (a + c) (b + c) is positive,\nMultiply both \ sides by 2 (a + b) (a + c) (b + c) to get:\n9 (a + b) (a + c) (b + c) > \n \n \ 3 2 2 3 2 2 2\n 2 (a + a b + a \ b + b + a c + 3 a b c + b c + a c + \n \n 2 3 2 \ 2\n b c + c + a s + 3 a b s + b s + 3 a c s + \n \n \ 2\n 3 b c s + c s)\nExpanding and collecting terms of the same sign \ gives:\n 2 2 2 2 2\n7 a b + 7 a b \ + 7 a c + 12 a b c + 7 b c + 7 a c + \n \n 2 3 3 3 \ 2\n 7 b c > 2 a + 2 b + 2 c + 2 a s + 6 a b s + \n \n 2 \ 2\n 2 b s + 6 a c s + 6 b c s + 2 c s\nLet s=(a + b \ + c)/2.\nWe get:\n 2 2 2 2 2\n7 a \ b + 7 a b + 7 a c + 12 a b c + 7 b c + 7 a c + \n \n 2 3 \ 3 3\n 7 b c > 2 a + 2 b + 2 c + \n \n 2 \ 2 2\n (a + b + c) (2 a + 6 a b + 2 b + 6 a c + 6 \ b c + 2 c )\n --------------------------------------------------------\n \ 2\nExpanding and collecting terms of the same sign \ gives:\n 2 2 2 2 2\n3 a b + 3 a b \ + 3 a c + 3 a b c + 3 b c + 3 a c + \n \n 2 3 3 3\n \ 3 b c > 3 a + 3 b + 3 c\nwe can divide both sides by 3 to get:\n 2 \ 2 2 2 2 2\na b + a b + a c + a b c + b c + a \ c + b c > \n \n 3 3 3\n a + b + c\nLet a=(y+z)/2, b=(z+x)/2, \ c=(x+y)/2, to get:\n 2 2 2\n(x + y) \ (x + z) (x + y) (x + z) (x + y) (y + z)\n---------------- + \ ---------------- + ---------------- + \n 8 8 \ 8\n \n 2\n (x + y) (x + z) (y + \ z) (x + z) (y + z)\n ----------------------- + ---------------- + \n \ 8 8\n \n 2 2\n \ (x + y) (y + z) (x + z) (y + z)\n ---------------- + ---------------- \ > \n 8 8\n \n 3 3 3\n (x \ + y) (x + z) (y + z)\n -------- + -------- + --------\n 8 \ 8 8\nMultiply both sides by 8 to get:\n 3 2 2 3 \ 2 2\n2 (x + 3 x y + 3 x y + y + 3 x z + 7 x y z + 3 \ y z + \n \n 2 2 3\n 3 x z + 3 y z + z ) > \n \n \ 3 2 2 3 2 2\n 2 x + 3 x y + 3 x y + 2 y + \ 3 x z + 3 y z + \n \n 2 2 3\n 3 x z + 3 y z + 2 z\n\ Expanding and collecting terms of the same sign gives:\n 2 2 \ 2 2 2\n3 x y + 3 x y + 3 x z + 14 x y z + 3 y \ z + 3 x z + \n \n 2\n 3 y z > 0\nThis is true because a sum of \ positive terms must be\\\n \n positive.\n \n 2 2 2\n \ a + b + c 1\nTo prove: ------------ >= -\n 2 3\n\ (a + b + c)\n 2\nSince 3 (a + b + c) is \ positive,\n 2\nMultiply both sides by 3 (a \ + b + c) to get:\n 2 2 2 2\n3 (a + b + c ) >= (a + \ b + c)\nExpanding and collecting terms of the same sign gives:\n 2 2 \ 2\n2 a + 2 b + 2 c >= 2 a b + 2 a c + 2 b c\nwe can divide both sides \ by 2 to get:\n 2 2 2\na + b + c >= a b + a c + b c\nLet a=(y+z)/2, \ b=(z+x)/2, c=(x+y)/2, to get:\n 2 2 2\n(x + y) (x \ + z) (y + z)\n-------- + -------- + -------- >= \n 4 4 \ 4\n \n (x + y) (x + z) (x + y) (y + z) (x + z) (y + z)\n \ --------------- + --------------- + ---------------\n 4 \ 4 4\nMultiply both sides by 4 to get:\n 2 2 \ 2\n2 (x + x y + y + x z + y z + z ) >= \n \n 2 2 \ 2\n x + 3 x y + y + 3 x z + 3 y z + z\nExpanding and \ collecting terms of the same sign gives:\n 2 2 2\nx + y + z >= x y + \ x z + y z\nExpressing in terms of symmetric polynomials gives:\n[2,0,0] >= \ [1,1,0]\nThis result follows from the majorization theorem.\n \n \ 2 2 2\n 1 a + b + c\nTo prove: - > ------------\n \ 2 2\n (a + b + c)\n 2\nSince 2 \ (a + b + c) is positive,\n 2\nMultiply \ both sides by 2 (a + b + c) to get:\n 2 2 2 2\n(a + b \ + c) > 2 (a + b + c )\nExpanding and collecting terms of the same sign \ gives:\n 2 2 2\n2 a b + 2 a c + 2 b c > a + b \ + c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n (x + y) (x + z) (x + \ y) (y + z) (x + z) (y + z)\n2 (--------------- + --------------- + \ ---------------) > \n 4 4 4\n \n \ 2 2 2\n (x + y) (x + z) (y + z)\n -------- + \ -------- + --------\n 4 4 4\nMultiply both sides by 2 \ to get:\n 2 2 2\nx + 3 x y + y + 3 x z + 3 y \ z + z > \n \n 2 2 2\n x + x y + y + x z + y z + \ z\nExpanding and collecting terms of the same sign gives:\n2 x y + 2 x z + 2 \ y z > 0\nThis is true because a sum of positive terms must be\\\n \n \ positive.\n \n 3 3 3 5 5 5\nTo prove: 2 (a + b \ + c ) (a + b + c ) > \n \n 2 2 2 3 3 3 3 3 3\n (a \ + b + c ) (a b + a c + b c )\nExpanding and collecting terms of the \ same sign gives:\n 8 5 3 3 5 8 5 3 5 3 3 5\n2 a + \ a b + a b + 2 b + a c + b c + a c + \n \n 3 5 8 3 3 \ 2 3 2 3 2 3 3\n b c + 2 c > a b c + a b c + a b c\nLet \ a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n 8 5 3 \ 3 5\n(x + y) (x + y) (x + z) (x + y) (x + z)\n-------- + \ ----------------- + ----------------- + \n 128 256 \ 256\n \n 8 5 3 5 3\n (x + z) \ (x + y) (y + z) (x + z) (y + z)\n -------- + ----------------- + \ ----------------- + \n 128 256 256\n \n \ 3 5 3 5 8\n (x + y) (y + z) (x + z) \ (y + z) (y + z)\n ----------------- + ----------------- + -------- > \n \ 256 256 128\n \n 3 3 \ 2 3 2 3\n (x + y) (x + z) (y + z) (x + y) (x + \ z) (y + z)\n -------------------------- + -------------------------- + \n \ 256 256\n \n 2 3 3\n\ (x + y) (x + z) (y + z)\n --------------------------\n \ 256\nMultiply both sides by 256 to get:\n 8 7 6 2 5 \ 3 4 4\n6 x + 24 x y + 69 x y + 125 x y + 150 x y + \n \n \ 3 5 2 6 7 8 7\n 125 x y + 69 x y + 24 x y \ + 6 y + 24 x z + \n \n 6 5 2 4 3 3 4\n \ 30 x y z + 51 x y z + 55 x y z + 55 x y z + \n \n 2 5 \ 6 7 6 2\n 51 x y z + 30 x y z + 24 y z + 69 x z + \n \ \n 5 2 4 2 2 3 3 2 2 4 2\n 51 x y z + \ 90 x y z + 100 x y z + 90 x y z + \n \n 5 2 6 2 \ 5 3 4 3\n 51 x y z + 69 y z + 125 x z + 55 x y z + \n \n\ 3 2 3 2 3 3 4 3 5 3\n 100 x y z + \ 100 x y z + 55 x y z + 125 y z + \n \n 4 4 3 4 \ 2 2 4 3 4\n 150 x z + 55 x y z + 90 x y z + 55 x y z + \ \n \n 4 4 3 5 2 5 2 5\n 150 y z + 125 x \ z + 51 x y z + 51 x y z + \n \n 3 5 2 6 6 \ 2 6 7\n 125 y z + 69 x z + 30 x y z + 69 y z + 24 x z + \n\ \n 7 8 6 2 5 3 4 4 3 5\n 24 y z + 6 z \ > x y + 5 x y + 8 x y + 5 x y + \n \n 2 6 6 5 2 \ 4 3\n x y + 2 x y z + 15 x y z + 35 x y z + \n \n 3 4 \ 2 5 6 6 2\n 35 x y z + 15 x y z + 2 x y z + x \ z + \n \n 5 2 4 2 2 3 3 2 2 4 2\n 15 x y \ z + 54 x y z + 80 x y z + 54 x y z + \n \n 5 2 6 2 \ 5 3 4 3\n 15 x y z + y z + 5 x z + 35 x y z + \n \n \ 3 2 3 2 3 3 4 3 5 3\n 80 x y z + 80 x y z + \ 35 x y z + 5 y z + \n \n 4 4 3 4 2 2 4 3 \ 4\n 8 x z + 35 x y z + 54 x y z + 35 x y z + \n \n 4 4 \ 3 5 2 5 2 5 3 5\n 8 y z + 5 x z + 15 x y z + \ 15 x y z + 5 y z + \n \n 2 6 6 2 6\n x z + 2 x y z \ + y z\nExpanding and collecting terms of the same sign gives:\n 8 7 \ 6 2 5 3 4 4\n6 x + 24 x y + 68 x y + 120 x y + \ 142 x y + \n \n 3 5 2 6 7 8 7\n 120 x \ y + 68 x y + 24 x y + 6 y + 24 x z + \n \n 6 5 2 \ 4 3 3 4\n 28 x y z + 36 x y z + 20 x y z + 20 x y z + \n \ \n 2 5 6 7 6 2\n 36 x y z + 28 x y z + \ 24 y z + 68 x z + \n \n 5 2 4 2 2 3 3 2 2 \ 4 2\n 36 x y z + 36 x y z + 20 x y z + 36 x y z + \n \n \ 5 2 6 2 5 3 4 3\n 36 x y z + 68 y z + 120 x z \ + 20 x y z + \n \n 3 2 3 2 3 3 4 3 5 3\n \ 20 x y z + 20 x y z + 20 x y z + 120 y z + \n \n 4 4 \ 3 4 2 2 4 3 4\n 142 x z + 20 x y z + 36 x y z + \ 20 x y z + \n \n 4 4 3 5 2 5 2 5\n 142 \ y z + 120 x z + 36 x y z + 36 x y z + \n \n 3 5 2 6 \ 6 2 6 7\n 120 y z + 68 x z + 28 x y z + 68 y z \ + 24 x z + \n \n 7 8\n 24 y z + 6 z > 0\nThis is true \ because a sum of positive terms must be\\\n \n positive.\n \nTo prove: -3 a \ b c + 5 (a b (a + b) + a c (a + c) + \n \n 3\ \n b c (b + c)) >= (a + b + c)\nExpanding and collecting terms of the \ same sign gives:\n 2 2 2 2 2 2\n2 a b \ + 2 a b + 2 a c + 2 b c + 2 a c + 2 b c >= \n \n 3 3 3\ \n a + b + 9 a b c + c\nLet a=(y+z)/2, b=(z+x)/2, c=(x+y)/2, to get:\n \ 2 2 2\n (x + y) (x + z) (x + y) \ (x + z) (x + y) (y + z)\n2 (---------------- + ---------------- + \ ---------------- + \n 8 8 8\n \n \ 2 2 2\n (x + z) (y + \ z) (x + y) (y + z) (x + z) (y + z)\n ---------------- + \ ---------------- + ----------------)\n 8 8 \ 8\n \n 3 3\n (x + y) (x + z) 9 (x \ + y) (x + z) (y + z)\n >= -------- + -------- + ------------------------- \ + \n 8 8 8\n \n 3\n (y + z)\n \ --------\n 8\nMultiply both sides by 4 to get:\n 3 2 2 \ 3 2\n2 x + 5 x y + 5 x y + 2 y + 5 x z + 12 x y z + \n \n \ 2 2 2 3\n 5 y z + 5 x z + 5 y z + 2 z >= \n \n 3 \ 2 2 3 2 2\n x + 6 x y + 6 x y + y \ + 6 x z + 9 x y z + 6 y z + \n \n 2 2 3\n 6 x z + 6 y z \ + z\nExpanding and collecting terms of the same sign gives:\n 3 3 \ 3\nx + y + 3 x y z + z >= \n \n 2 2 2 2 2 \ 2\n x y + x y + x z + y z + x z + y z\nExpressing in terms of \ symmetric polynomials gives:\n[1,1,1] + [3,0,0] >= 2 [2,1,0]\nThis follows \ from the following majorizations:\n[1,1,1] + [3,0,0] >= 2 [2,1,0]\nSummary:\n\ Proved 23 inequalities out of 23."], "Print", Evaluatable->False, AspectRatioFixed->True]}, Open]] }, FrontEndVersion->"Macintosh 3.0", ScreenRectangle->{{0, 1152}, {0, 850}}, AutoGeneratedPackage->None, WindowToolbars->{}, CellGrouping->Manual, WindowSize->{520, 740}, WindowMargins->{{68, Automatic}, {Automatic, 28}}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, -1}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, MacintoshSystemPageSetup->"\<\ 02P0001804P000000^L2D_ogooL33`9K8085:0?l0000005X0FP000003X<;VP5d 038;VTRX04/00@4100000BL?00400@000000000000000000000000000040I0<0 00000000002@X@@08@000000000e@b/P\>" ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. ***********************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1731, 51, 1436, 23, 70, "Input", InitializationCell->True], Cell[3170, 76, 57472, 733, 70, "Print", Evaluatable->False] }, Open ]] } ] *) (*********************************************************************** End of Mathematica Notebook file. ***********************************************************************)