Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial derivatives and finding the maximum, is there a simple argument that imply this inequality (perhaps using symmetry somehow?).
Thanks !
 A: $$\frac{(x+y)^2}{x^2+xy+y^2} = 1 + \frac{xy}{x^2+xy+y^2}$$
$$xy \le \frac{(x^2 + y^2)}{2} $$ 
$$\therefore \frac{3xy}{2} \le \frac{(x^2 + y^2 + xy)}{2}$$
$$\therefore \frac{xy}{x^2+xy+y^2} \le \frac 13$$
$$\therefore \frac{(x+y)^2}{x^2+xy+y^2} \le \frac 43 $$
A: We have $$\frac{(x+y)^2}{x^2+xy+y^2} = \frac{(x+y)^2}{\frac{3}{4}(x+y)^2 + \frac{1}{4}(x-y)^2}\leq \frac{(x+y)^2}{\frac{3}{4}(x+y)^2}=\frac{4}{3}$$
The inequality holds since squares are non-negative.
A: The reciprocal is
$$\frac{x^2+xy+y^2}{(x+y)^2}=1-\frac{xy}{(x+y)^2} $$
and by the AMGM inequality, $\sqrt{xy}\le \frac{x+y}2$ with equality iff $x=y$, hence $\frac{xy}{(x+y)^2}\le \frac14$ with equality iff $x=y$ and from this  $\frac{x^2+xy+y^2}{(x+y)^2}\ge \frac 34$ and finally $\frac{(x+y)^2}{x^2+xy+y^2}\le \frac 43$
A: $$\frac{(x+y)^2}{x^2+xy+y^2}=1+\frac{xy}{x^2+xy+y^2}\le 1+\frac{1}{3}$$  
$$\iff 3xy\le x^2+xy+y^2\iff (x-y)^2\ge 0,$$
which is true, with equality iff $x=y$.   
Note we could multiply by $x^2+xy+y^2$ without flipping ineq sign because it equals $\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2>0$
A: Since $x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac{3y^2}{4}\gt 0$, one has$$\begin{align}\frac{(x+y)^2}{x^2+xy+y^2}\le\frac 43&\iff (x+y)^2\le\frac{4}{3}(x^2+xy+y^2)\\&\iff \frac 13x^2-\frac 23xy+\frac 13y^2\ge0\\&\iff \frac 13(x-y)^2\ge 0\end{align}$$
This is true.
