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 !

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.
