<s>[INST] inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$ I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:
$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$
It's rather easy to show
$$
\frac{4xy}{x^2 + y^2} \le 2
$$
and obviously
$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$
but combining these two does not give me a sufficient low bound.
 [/INST] How about this method?
$$\frac{2(x + y)²}{2x² + y²}-3=\frac{2x^2+4xy+2y^2-(6x^2+3y^2)}{2x^2+y^2}=-\frac{4x^2+4xy+y^2}{2x^2+y^2}=-\frac{(2x-y)^2}{2x^2+y^2}\le0..$$
</s>