# An Inequality in a Nonobtuse Triangle

### Solution

Let $a=\sqrt{y+z},$ $b=\sqrt{z+x},$ $c=\sqrt{x+y},$ with $x\ge y\ge z\ge 0,$ $y\gt 0.$ In terms of $x,y,z$ the inequality to prove becomes

$2(xy+yz+zx)^2-(y+z)^2(x+y)(x+z)\ge 0.$

Clearly, the quadratic function $f(x)=2(xy+yz+zx)^2-(y+z)^2(x+y)(x+z)$ is strictly increasing on $[y,\infty),$ so that $f(x)\ge f(y)=2y^2(y+2x)^2-2y(y+z)^3.$

Obviously, $2y^2(y+2x)^2-2y(y+z)^3\ge 0$ with equality iff $z=0,$ i.e. if $\Delta ABC$ is right isosceles and the altitude in question is its smaller side.

### Acknowledgment

Leo Giugiuc has kindly communicated to me the problem and a solution of his. The problem is by Alijadallah Belabess.

[an error occurred while processing this directive]