TITLE: Showing an inequality: $\sqrt{xy} \leq \frac{2xy}{x+y}$
QUESTION [5 upvotes]: For $x,y \in \mathbb{R}_{0}^+$ I want to show that
$$\sqrt{xy} \leq \frac{2xy}{x+y}$$
only if $x=y$. I think squaring both sides is a equivalence transformation due to $x>0, y>0$ but it didn't get me anywhere.
Can someone give me a hint or a solution?

REPLY [2 votes]: Since $x$ and $y$ are positive, you can write $u=\sqrt{x}$ and $v=\sqrt{y}$, so the inequality is
$$
uv\le\frac{2u^2v^2}{u^2+v^2}
$$
that's equivalent to
$$
u^2+v^2\le 2uv
$$
or
$$
(u-v)^2\le 0
$$