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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!

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

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Flavio Esposito
    Jan 8, 2020 at 1:07
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    $\begingroup$ 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. $\endgroup$
    – user317176
    Jan 8, 2020 at 1:11
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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

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

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