Minimum of $\frac{x}{1+y^2}+\frac{y}{1+x^2}$ on $x,y\ge 0$, $x+y=2$

Question

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?

Answer 1

Hint:

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

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Addendum:

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

Answer 2

y=2-x.

Then plug it in the expression. Then use derivatives.

Answer 3

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} $

Answer 4

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 $x=y=1$