<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] $$\frac{(x+y)^2}{x^2+xy+y^2} = 1 + \frac{xy}{x^2+xy+y^2}$$
$$xy \le \frac{(x^2 + y^2)}{2} $$ 
$$\therefore \frac{3xy}{2} \le \frac{(x^2 + y^2 + xy)}{2}$$
$$\therefore \frac{xy}{x^2+xy+y^2} \le \frac 13$$
$$\therefore \frac{(x+y)^2}{x^2+xy+y^2} \le \frac 43 $$
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