<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] Let $y=l.x$; we have:
$\frac{2[(l+1)x]^2}{x^2(l^2+2)}=\frac {2l^2+4l+2}{l62+2}=2+\frac{4l-2}{l^2+2}$
$\frac{4l-2}{l^2+2}<1$
because:
$(l-2)^2>0$
$l^2-4l+2+2>0$
$4l-2<l^2+2$
If $l=2$ the equality holds.
</s>