Problem
ALG-B2-M08-P019 Double triangle denominators
#19
★★★★★ Level 5 of 5
For triangle sides, prove \[\frac{a^2}{(b+c-a)(c+a-b)}+\frac{b^2}{(c+a-b)(a+b-c)}+\frac{c^2}{(a+b-c)(b+c-a)}\ge3.\]
Hint. Use \(a=y+z\), \(b=z+x\), \(c=x+y\), then bound each fraction by one ratio.
Then \(b+c-a=2x\), \(c+a-b=2y\), \(a+b-c=2z\). The first fraction is \(\frac{(y+z)^2}{4xy}\ge\frac{z}{x}\), because \((y+z)^2\ge4yz\). Similarly, the second fraction is at least \(\frac{x}{y}\), and the third at least \(\frac{y}{z}\). Hence the whole sum is at least \(\frac{z}{x}+\frac{x}{y}+\frac{y}{z}\ge3\).
Teaching goal: recognize the appropriate substitution and check the domain.