<s>[INST] 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
$$\left|\frac{x+y}{2}+\sqrt{xy}\right| = \left|\frac{ (\sqrt{x}+\sqrt{y})^2}{2}\right|$$
 [/INST] Hint: It may be clearer if you write $x=(-1)^t2a^2$, $y=(-1)^t2b^2$ for some $a,b>0$ and $t\in\{0,1\}$.
Mouse over the gray box for more details if you get stuck.

 You now want to show that $$|(-1)^t(a^2+b^2)-2ab|+|(-1)^t(a^2+b^2)+2ab|=2a^2+2b^2.$$ Shuffling the minus signs, we get $$|a^2+b^2+(-1)^{t+1}2ab|+|a^2+b^2+(-1)^t2ab|=2a^2+2b^2.$$ Whether or not $t=0$ or $t=1$, one of the two terms on the left is equal to $$|a^2+b^2-2ab|=|(a-b)^2|=(a-b)^2=a^2+b^2-2ab,$$ and the other term is equal to $$|a^2+b^2+2ab|=|(a+b)^2|=(a+b)^2=a^2+b^2+2ab.$$

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