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 !

$$\frac{(x+y)^2}{x^2+xy+y^2}=1+\frac{xy}{x^2+xy+y^2}\le 1+\frac{1}{3}$$  
$$\iff 3xy\le x^2+xy+y^2\iff (x-y)^2\ge 0,$$
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 $\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2>0$
