# An Inequality in Triangle form the 1996 APMO

### Problem

### Solution 1

Set $\displaystyle s={a+b+c}{2},$ and $x=s-a,$ $y=s-b,$ $z=s-c$ (that is known as *Ravi's substitution*.) The required inequality reduces to

(*)

$\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{y+z}+\sqrt{z+x}+\sqrt{x+y}.$

Recall now the AM-QM inequality, $\displaystyle \frac{u+v}{2}\le\sqrt{\frac{u^2+v^2}{2}}$ which, for, say, $u=\sqrt{2x}$ and $v=\sqrt{2y}$ gives $\displaystyle \frac{\sqrt{2x}+\sqrt{2y}}{2}\le\sqrt{\frac{2x+2y}{2}}=\sqrt{x+y}.$ This confirms (*). Equality holds only for $x=y=z,$ or, which is the same, for $a=b=c.$

### Solution 2

Using Jensen's inequality,

$\displaystyle \sqrt{a}=\sqrt{\frac{(a+b-c)}{2}+\frac{(c+a-b)}{2}} \geq \frac{1}{2}\left(\sqrt{a+b-c}+\sqrt{c+a-b}\right).$

We get the desired inequality by collecting the other two cyclically generated inequalities and summing.

### Acknowledgment

This is problem 5 from the 1996 APMO. It was published in the Canadian Crux Mathematicorum (v25 n2) as Problem of the Month with a solution (Solution 1) by Jimmy Chui. Solution 2 is by Amit Itagi.

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