Question by Kayasari Ryuunosuke

28/07/2017 at 14:50

Supports a,b,c,d are non-negetive numbers so that ab + bc + cd + da = 1

Prove that : \(\dfrac{a^3}{b+c+d}+\dfrac{b^3}{a+c+d}+\dfrac{c^3}{a+b+d}+\dfrac{d^3}{a+b+c}\ge\dfrac{1}{3}\)

Lê Quốc Trần Anh Coordinator 29/07/2017 at 09:35

Trung Nguyễn, do you copy the solution in the link?
Lê Quốc Trần Anh Coordinator 28/07/2017 at 14:59

You can go to this link for the answer: https://mks.mff.cuni.cz/kalva/short/soln/sh9023.html
Trung Nguyễn 29/07/2017 at 09:15

Put s = w + x + y + z. Put A = w3/(s-w) + x3/(s-x) + y3/(s-y) + z3/(s-z), B = w2 + x2 + y2 + z2, C = w(s-w) + x(s-x) + y(s-y) + z(s-z) = 2 + 2wy + 2xz. By Cauchy-Schwartz, we have AC >= B2.

We have (w-x)2 + (x-y)2 + (y-z)2 + (z-w)2 ≥ 0, so B ≥ (wx + xy + yz + zw) = 1. Also (w - y)2 + (x - z)2 ≥ 0, so B ≥ 2wy + 2xz. So if 2wy + 2xz ≤ 1, then A ≥ 1/C ≥ 1/3. If 2wy + 2xz > 1, then C > 3, so (C-2)/C > 1 - 2/C > 1/3. Hence AC ≥ B2 ≥ B ≥ (C-2), so A > 1/3.