How can I Prove $\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$ for $x,y>0$ prove that
$\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$
I have tried to develop $(x+y)^2=$ and to get to an expression that must be bigger than those above
Thanks!
 A: $\sqrt{xy} \le \dfrac{x + y}{2}$ is just the AM–GM inequality. $\dfrac{2xy}{x + y} \le \sqrt{xy}$ is also the AM-GM inequality applied to $1/x$, $1/y$.
A: For example
$$\frac{2xy}{x+y}\le\sqrt{xy}\iff 4x^2y^2\le xy(x^2+2xy+y^2)\iff$$
$$ \iff xy(x^2+2xy+y^2-4xy)\ge 0\iff xy(x-y)^2\ge 0$$
And since the last rightmost inequality is obvious we're done.
A: For the 1st one , $(x+y)/2 \ge \sqrt{xy}$
$x+y\ge 2*\sqrt{xy}$
$(\sqrt{x})^2+(\sqrt{y})^2-2*\sqrt{xy}\ge0$
$(\sqrt{x}-\sqrt{y})^2\ge0$  ---->>> DONE
For the 2nd one :
$x+y\ge 2xy/\sqrt{xy}$
$x+y\ge 2*\sqrt{xy}$ --> this is the one just proved above by divide both side by 2.
A: Start by noticing that since $(x-y)^2 \geq 0$ then $x^2 - 2xy +y^2 \geq 0$ which implies that $x^2+y^2 \geq 2xy$. Now we attack these inequalities one at a time. 
To show the left inequality, we can show $2xy \leq (x+y)\sqrt{xy}$, and since everything is positive we can square both sides and further reduce the problem to showing $4x^2 y^2 \leq(x^2+2xy+y^2)(xy)$. Then, since $x^2+y^2 \geq 2xy$ we have $(x^2+2xy+y^2)(xy) \geq 4xy(xy) = 4x^2y^2$, which is exactly what you wanted to show.
To show the right inequality we will square both sides again and reduce the problem to showing $xy \leq \frac{1}{4}(x^2 + 2xy +y^2)$. To do so we again use that $x^2+y^2 \geq 2xy$. Therefore, $\frac{1}{4}(x^2+2xy+y^2) \geq \frac{1}{4}(4xy) = xy$ which is again the inequality you wanted to show.
