Problem
ALG-B2-M01-P006 Cyclic sum of fractions
#6
★★★★☆ Level 4 of 5
Let \(x,y,z>0\). Prove that \[\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>1.\]
Hint 1. Choose the largest of \(x,y,z\).
Hint 2. If \(x\ge z\), compare \(\frac{y}{y+z}\) and \(\frac{y}{y+x}\).
Cyclically rename the variables so that \(x\ge z\). Then \(y+z\le y+x\), hence \(\frac{y}{y+z}\ge\frac{y}{y+x}\). Therefore \[\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge\frac{x}{x+y}+\frac{y}{y+x}+\frac{z}{z+x}=1+\frac{z}{z+x}>1.\]
Module training problem. Method tags: ordering, fractions, bounds.