Problem
ALG-B2-M10-P009 Triangle fraction
#9
★★★★★ Level 5 of 5
Let \(a,b,c\) be triangle sides. Prove \[\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge3.\]
Hint. Use \(a=y+z\), \(b=z+x\), \(c=x+y\).
Then \(b+c-a=2x\), \(c+a-b=2y\), \(a+b-c=2z\). The sum is \(\frac12\sum\frac{y+z}{x}\). Inside are three pairs of the form \(\frac{y}{x}+\frac{x}{y}\), each at least \(2\), so the sum is at least \(3\).
Teaching goal: choose a combination of methods and always check the equality case.