MathDB
Inequality with xy+yz+zx=1

Source: Baltic Way 2022, Problem 4

November 12, 2022
inequalities

Problem Statement

The positive real numbers x,y,zx,y,z satisfy xy+yz+zx=1xy+yz+zx=1. Prove that: 2(x2+y2+z2)+43(1x2+1+1y2+1+1z2+1)5 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5