Problem
ALG-B2-M08-P014 Strong triangle fraction
#14
★★★★★ Level 5 of 5
For triangle sides, prove \[\frac{b+c}{b+c-a}+\frac{c+a}{c+a-b}+\frac{a+b}{a+b-c}\ge6.\]
Hint. After triangle substitution, the first fraction becomes \(\frac{2x+y+z}{2x}\).
Let \(a=y+z\), \(b=z+x\), \(c=x+y\). The sum becomes \(\sum\left(1+\frac{y+z}{2x}\right)=3+\frac12\sum\frac{y+z}{x}\). As in the example, \(\sum\frac{y+z}{x}\ge6\). Hence the sum is at least \(6\).
Teaching goal: recognize the appropriate substitution and check the domain.