Absolute value inequality - Please guide further 
Prove that if the numbers $x$, $y$ are of one sign, then $\left|\frac{x+y}{2}-\sqrt{xy}\right|+\left|\frac{x+y}{2}+\sqrt{xy}\right|=|x|+|y|$.

Expanding the LHS,
$$\left|\frac{x+y}{2}-\sqrt{xy}\right|= \left|\frac{x+y -2\sqrt{xy}}{2}\right| = \frac{(\sqrt{x}-\sqrt{y})^2}{2}$$
and similarly
$$\left|\frac{x+y}{2}+\sqrt{xy}\right| = \left|\frac{ (\sqrt{x}+\sqrt{y})^2}{2}\right|$$

Use AM-GM to conclude $\frac{x+y}{2} \ge \sqrt{xy}$ if $x,y > 0$ and $\frac{x+y}{2} \le -\sqrt{xy}$ if $x,y < 0.$
