<s>[INST] 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?
 [/INST] 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}
$
</s>