MathDB
Inequality from 2017 Taiwan TST

Source: 2017 Taiwan TST Round 1

April 13, 2018
inequalities

Problem Statement

Given a,b,c,d>0a,b,c,d>0, prove that: cycca+2b+cyca+2bc8((a+b+c+d)2ab+ac+ad+bc+bd+cd1),\sum_{cyc}\frac{c}{a+2b}+\sum_{cyc}\frac{a+2b}{c}\geq 8(\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}-1), where cycf(a,b,c,d)=f(a,b,c,d)+f(d,a,b,c)+f(c,d,a,b)+f(b,c,d,a)\sum_{cyc}f(a,b,c,d)=f(a,b,c,d)+f(d,a,b,c)+f(c,d,a,b)+f(b,c,d,a).