# Proving an inequality using cases and the AM-GM inequality

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!

• What happens if you only square once and assume $x \ge y \ge 0$? Jan 8, 2020 at 0:36
• You could drop the absolute signs? Jan 8, 2020 at 0:40
• Indeed you could Jan 8, 2020 at 0:41

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.

• Thank you! So, your objective was to basically tweak the equation in such a way so it will resemble the AM-GM inequality, which allowed you to make use of it? Jan 8, 2020 at 1:07
• I had an answer that used the hint provided, but I deleted that portion of it. That did not use AM-GM. But, finishing the approach you started, my first goal was to simplify, and then It became more obvious that I could use AM-GM. Jan 8, 2020 at 1:11

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

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