Problem
ALG-B2-M08-P006 Triangle denominators
#6
★★★☆☆ Level 3 of 5
For triangle sides, prove \[\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge3.\]
Hint. Under the triangle substitution, \(b+c-a=2x\).
Let \(a=y+z\), \(b=z+x\), \(c=x+y\). The sum becomes \(\frac12\left(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\right)\). Inside are the pairs \(\frac{y}{x}+\frac{x}{y}\), \(\frac{z}{x}+\frac{x}{z}\), \(\frac{z}{y}+\frac{y}{z}\), each at least \(2\). Thus the sum is at least \(3\).
Teaching goal: recognize the appropriate substitution and check the domain.