<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] Well $2x^2+y^2> 0$ and $2(x+y)^2\ge 0$ so
So $\frac{2(x + y)^2}{2x^2+ y^2} \leq 3\iff$
$2(x+y)^2 \leq3(2x^2 + y^2) \iff$
$2x^2 + 4xy +2y^2 \leq 6x^2 + 3y^2 \iff$
$0 \leq 4x^2 -4xy +y^2 \iff$
$0 \leq(2x -y)^2 \iff$ God's in his heaven and all's right in the world.
</s>