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 !

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 $\frac{(x+y)^2}{x^2+xy+y^2}\le \frac 43$
