XXVIII-th Balkan Mathematical Olympiad

The inequality is clear if $xyz=0$, in which case equality holds if and only if $x=y=z=0$.

Henceforth assume $xyz\ne 0$ and rewrite the inequality as $$\frac{(2x+1{)}^2}{2x^2+1}+\frac{(2y+1{)}^2}{2y^2+1}+\frac{(2z+1{)}^2}{2z^2+1}\ge 3.$$ Notice that (exactly) one of the products $xy,yz,zx$ is positive, say $yz>0$, to get $$\begin{array}{rlrl}\frac{(2y+1{)}^2}{2y^2+1}+\frac{(2z+1{)}^2}{2z^2+1}& \ge \frac{2(y+z+1{)}^2}{y^2+z^2+1}& & \text{(by Jensen)}\\ & =\frac{2(x-1{)}^2}{x^2-2yz+1}& & \text{(for x+y+z=0)}\\ & \ge \frac{2(x-1{)}^2}{x^2+1}& & \text{(for yz>0)}\end{array}$$ Here equality holds if and only if $x=1$ and $y=z=-\frac{1}{2}$.

Finally, since $$\frac{(2x+1{)}^2}{2x^2+1}+\frac{2(x-1{)}^2}{x^2+1}-3=\frac{2x^2(x-1{)}^2}{(2x^2+1)(x^2+1)}\ge 0,x\in \mathbb{R},$$ The conclusion follows. Clearly, equality holds if and only if $x=1$, so $y=z=-\frac{1}{2}$. Therefore, if $xyz\ne 0$, equality holds if and only if one of the numbers is $1$, and the other two are $-\frac{1}{2}$.