For $x,y$ strictly positive is it true that $\frac{xy}{(x+y)^2}$ is bounded by $1/4$

Question

For $x$ and $y$ strictly positive is it true that $\frac{xy}{(x+y)^2}$ is bounded by $1/4$, or more formally is $$\sup_{(x,y)\in(0,\infty)\times (0,\infty)} \frac{xy}{(x+y)^2}\leq \frac{1}{4}.$$ I lag knowledge about functions of more variables so I'm really uncertain if the below argument holds, can someone help me?

My attempt for an argument goes as follows. For any point $(x',y')$ define the function $f(x)=\frac{x y'}{(x+y')^2}$ which has first derivative $\frac{y'(y'-x)}{(x+y')^3}$, hence increasing for $x$ less than $y'$ and decreasing for $x$ greater than $y'$, so $x^*=y$ is a unique maxima for $f$, hence $\frac{x'y'}{(x'+y')^2}=f(x')\leq f(x^*)=f(y')=\frac{y'y'}{(y'+y')^2}=\frac{1}{4}$, which shows that $\frac{1}{4}$ is an upperbound for any point(since $\frac{1}{4}$ is also atained by a point i guess it will also be eqaul to the supremum if the above argument is correct).

Answer 1

According to the AM-GM inequality, it results that \begin{align*} x + y \geq 2\sqrt{xy} \Rightarrow (x+y)^{2}\geq 4xy \Rightarrow \frac{xy}{(x+y)^{2}} \leq \frac{1}{4} \end{align*}

Answer 2

$$\frac{xy}{(x+y)^2}\le 1/4$$

$$(x+y)^2\ge 4xy$$ $$ x^2+y^2\ge 2xy$$ $$ (x-y)^2\ge 0$$

Answer 3

For any $a>0$, by rearrangement inequality, we have $a+\frac1a\ge 2$ therefore

$$\frac{xy}{(x+y)^2}=\frac{1}{\left(\frac {\sqrt x} {\sqrt y}+\frac {\sqrt y} {\sqrt x}\right)^2}\le \frac 1 4$$

Refer also to the related

• How to prove $\frac xy + \frac yx \ge 2$

Answer 4

$$(x+y)^2-(x-y)^2=4xy$$ $$ (x+y)^2\ge 4xy$$ $$\frac {xy}{(x+y)^2}\le 1/4$$