Showing an inequality: $\sqrt{xy} \leq \frac{2xy}{x+y}$ 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?
 A: Both $x$ and $y$ are assumed to be positive, so you can
square the inequality. This gives
$$
xy \le \frac {4x^2y^2}{(x+y)^2} \\
\Longleftrightarrow (x+y)^2 \le 4 xy \\
\Longleftrightarrow (x -y)^2 \le 0 
$$
which apparently holds only for $x = y$.
A: After taking reciprocals the inequality is equivalent to 
$$\sqrt{\frac1x\frac1y}\ge \frac{1/x+1/y}2 $$
whereas the arithmetic-geometric inequality for $\frac1x$ and $\frac 1y$ states 
$$\sqrt{\frac1x\frac1y}\le \frac{1/x+1/y}2 $$
with equality iff $\frac1x=\frac1y$
A: 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
$$
(u-v)^2\le 0
$$
A: In general, to prove these types of inequalities you want to work backward (i.e., simplify what you're given) until you get something you know is true, then work forward for your proof. For example, getting rid of the fraction and squaring both sides, then combining terms and factoring gives something true...
Notice that 
$$(x-y)^2\geq 0.$$
This gives
$$0\leq x^2-2xy+y^2,$$
add $4xy$ to both sides,
$$4xy\leq x^2+2xy+y^2=(x+y)^2,$$
multiply both sides by $xy$ to get
$$(2xy)^2\leq xy(x+y)^2$$
then divide $(x+y)^2$ over and take the square root to get
$$\sqrt{xy}\geq\frac{2xy}{x+y}.$$
Of course equality holds when $x=y$.
