Minimum of $\frac{x}{1+y^2}+\frac{y}{1+x^2}$ on $x,y\ge 0$, $x+y=2$ let $x,y\ge 0$, and such $x+y=2$  find the  minimum 
$$\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}$$
I think $x=y=1$ is minimum of the value $1$,How can I prove?

Hint:
The minimum is indeed when $x=y=1$. Using AM-GM or Cauchy Schwarz inequality, it is sufficient to show $xy^2+yx^2\le 2$.

Addendum:  

$$\begin{align}\frac{x}{1+y^2}+\frac{x(1+y^2)}4 \ge x, &\quad \frac{y}{1+x^2}+\frac{y(1+x^2)}4 \ge y \\ \implies \frac{x}{1+y^2}+\frac{y}{1+x^2} &\ge \frac32-\frac{xy^2+yx^2}4 \\&= \frac32-\frac{x+y}4xy \\&\ge \frac32-\frac{x+y}4\frac{(x+y)^2}4=1\end{align} $$

