Problem
ALG-B2-M11-P006 Squares in a cycle
#6
★★★★☆ Level 4 of 5
Prove for \(a,b,c>0\): \[\frac{a^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+a^2}\ge1.\]
Hint. Put \(x=a^2\), \(y=b^2\), \(z=c^2\), and bring \(\sum x/(x+y)-1\) to a common denominator.
It is enough to prove for \(x,y,z>0\) that \(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge1\). With common denominator \((x+y)(y+z)(z+x)\), the numerator of the difference is \(xyz+x^2y+xz^2+y^2z\), which is positive. Now take \(x=a^2\), \(y=b^2\), \(z=c^2\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.