<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] 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 $x=y=1$
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