Given real numbers $x,y,z$ such that $ x+y+z=0$, show that
\[ \frac{x(x+2)}{2x^2+1}+\frac{y(y+2)}{2y^2+1}+\frac{z(z+2)}{2z^2+1}\ge 0 \]

proof:(Vo Quoc Ba Can)

Notice that
\[\frac{x(x+2)}{2x^2+1}=\frac{(2x+1)^2}{2(2x^2+1)}-\frac{1}{2}.\]
So the inequality can be written as
\[\sum \frac{(2x+1)^2}{2x^2+1} \ge 3.\]
Now, using the Cauchy-Schwarz inequality, we see that
\[2x^2=\frac{4}{3}x^2+\frac{2}{3}(y+z)^2 \le \frac{4}{3}x^2+\frac{4}{3}(y^2+z^2).\]
Therefore,
\[ \sum \frac{(2x+1)^2}{2x^2+1} \ge 3\sum \frac{(2x+1)^2}{4(x^2+y^2+z^2)+3}=3.\]