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?
 A: 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:  

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

A: y=2-x.
Then plug it in the expression.
Then use derivatives.
A: Since it seems that
we get the minimum when
$x = y$,
I will assume that
$x \ne y$
and see what happens.
When $x = y = 1$,
the value of
$\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}
$
is
$1$.
Then,
if $x+y = 2$
and $x \ne y$,
$\begin{array}\\
\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}-1
&=\frac{x(1+x^2)+y(1+y^2)-(1+x^2)(1+y^2)}{(1+x^2)(1+y^2)}\\
&=\frac{x+x^3+y+y^3-(1+x^2+y^2+x^2y^2)}{(1+x^2)(1+y^2)}\\
&=\frac{x+y+x^3+y^3-(1+x^2+y^2+x^2y^2)}{(1+x^2)(1+y^2)}\\
&=\frac{2+2(x^2-xy+y^2)-(1+x^2+y^2+x^2y^2)}{(1+x^2)(1+y^2)}
\qquad\text{since }x+y = 2\\
&=\frac{1+x^2-2xy+y^2-x^2y^2}{(1+x^2)(1+y^2)}\\
&=\frac{1-x^2y^2+(x-y)^2}{(1+x^2)(1+y^2)}\\
&\gt 0
\qquad\text{since } x \ne y
\text{ and }xy < ((x+y)/2)^2 = 1\\
\end{array}
$
A: There is yet another way of proving this inequality.
$\dfrac{x}{1+y^2} =x(1-\dfrac{y^2}{1+y^2})=x-\dfrac{xy^2}{1+y^2} \geq x-\dfrac{xy^2}{2y}=x-\dfrac{xy}{2}$
Similarly, $\dfrac{y}{1+x^2} \geq y-\dfrac{xy}{2}$
Adding these up, we have $\dfrac{x}{1+y^2} + \dfrac{y}{1+x^2} \geq (x+y) -xy = 2-xy$
Also, $xy \leq \dfrac{1}{4} (x+y)^2=1 \Rightarrow -xy \geq -1$
Therefore, $\dfrac{x}{1+y^2} + \dfrac{y}{1+x^2} \geq 2-1=1$
Finally, it can be seen that the equality indeed holds when $x=y=1$
