Home > IMO, Inequalities, Olympiad > Balkan Mathematical Olympiad 2011 Problem 2

## Balkan Mathematical Olympiad 2011 Problem 2

Given real numbers $x,y,z$ such that $x+y+z=0$, show that
$\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0$
When does equality hold?

BMO 2011 Problem 2

Categories: IMO, Inequalities, Olympiad Tags: , ,
1. May 18, 2011 at 7:10 am

$\frac{x(x+2)}{2x^2+1} +\frac{1}{2}=\frac{(2x+1)^2}{2(2x^2+1)}$.

$\sum_{cyc}\frac{(2x+1)^2}{2x^2+1}\ge 3$.

Assuming $xy \ge 0$ we have

$\sum_{cyc}\frac{(2x+1)^2}{2x^2+1} \ge \frac{(2x+2y+2)^2}{2(x^2+y^2)+2}+\frac{(2z+1)^2}{2z^2+1}\ge \frac{2(1-z)^2}{z^2+1}+\frac{(2z+1)^2}{2z^2+1} =3 + \frac{2z^2(z-1)^2}{(z^2+1)(2z^2+1)}\ge 3$.

• May 18, 2011 at 8:13 am

Nice solution. Thanks. 🙂