<s>[INST] 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 !
 [/INST] 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.
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