Proving an inequality using cases and the AM-GM inequality

Question

The question is the following:

I wanted to solve this question using the AM-GM inequality first and then trying to prove using cases. However, I am not exactly sure how to proceed.

What I have done so far (for the proof involving AM-GM):

I tried squaring both sides of the inequality (twice) to somehow make a use of the AM-GM inequality for two variables ($\frac{a+b}{2} \ge \sqrt{ab}$), and that is what I have tried doing: $$(x-2\sqrt{xy}+y)^2 \leq (|x-y|)^2$$ $$x^2+6xy+y^2-4x\sqrt{xy}-4y\sqrt{xy} \leq x-y$$

However, this didn't really help me much in finding the desired values of $a$ and $b$ to make the AM-GM inequality work. Could someone perhaps point me to the right direction?

Thanks!

Answer 1

The RHS is not multiplied out correctly

Squaring both sides twice.

$(x + y - 2\sqrt {xy})^2 \le (x-y)^2\\ x^2 + y^2 + 6xy - 4x\sqrt {xy} - 4y\sqrt {xy} \le x^2 - 2xy + y^2$

$8xy \le 4(x+y)\sqrt {xy}\\ (xy)^\frac 12 \le \frac 12 (x+y)$

Which is AM-GM

Or you could use the hint, and only square both sides once.

Answer 2

A bit late answer but I think it is worth mentioning it.

The inequality reminds me of the reverse triangle inequality $||a|-|b|| \leq |a-b|$, which gives an idea how to prove your inequality without cases and without AM-GM.

You only need the fact

• $(\star)$: For $a,b \geq 0$ we have $\sqrt{a+b}\leq \sqrt a + \sqrt b$ (which follows immediately by squaring)

Now, using this we have

$$\sqrt x = \sqrt{x-y+y} \stackrel{(\star)}{\leq} \sqrt{|x-y|}+\sqrt y \Leftrightarrow \boxed{\sqrt x - \sqrt y \leq \sqrt{|x-y|}}$$

and replacing $x$ and $y$

$$\sqrt y = \sqrt{y-x+x} \stackrel{(\star)}{\leq} \sqrt{|y-x|}+\sqrt x \Leftrightarrow \boxed{\sqrt y - \sqrt x \leq \sqrt{|x-y|}}$$

The two inequalities in the boxes give the required result

$$\left| \sqrt x - \sqrt y\right| \leq \sqrt{|x-y|}$$