Proving an inequality in three variables

Sam 05/15/2018. 3 answers, 196 views
real-analysis inequality

I thought the following inequality, which is to be proved, would be a simple application of the well known inequalities $||x|-|y|| \leq |x-y|$ and $||x|-|y|| \leq |x+y|$ but, if it is, after about an hour of scribbling I really can't seem to make any progress at all. Below is the inequality to be proved, for $x, y, z, \in \mathbb{R}$

$||x|-|y+z|| \leq ||x|-|y||+|z|$

For those interested, the problem is taken from page 5 of Mathematical Analysis: Functions of One Variable by M. Giaquinta and G. Modica

3 Answers


carmichael561 05/15/2018.

First, by adding and subtracting $|y|$ and using the triangle inequality, we obtain $$ ||x|-|y+z||=||x|-|y|+|y|-|y+z||\leq ||x|-|y||+||y|-|y+z||$$ and then using the inequality $$||a|-|b||\leq |a-b|$$ with $a=y$ and $b=y+z$, we get $$||x|-|y||+||y|-|y+z||\leq ||x|-|y||+|z|$$


mechanodroid 05/15/2018.

We have

$$|x|-\underbrace{|y+z|}_{\ge \big||y|-|z|\big|} \le |x|-\underbrace{\big||y|-|z|\big|}_{\ge |y| - |z|} \le \underbrace{|x| - |y|}_{\le \big||x| - |y|\big|} + |z| \le \big||x| - |y|\big| + |z|$$

On the other hand

$$|x|-\underbrace{|y+z|}_{\le |y| + |z|} \ge \underbrace{|x|- |y|}_{\ge -\big||x|- |y|\big|}-|z|\ge - \big||x|- |y|\big| - |z|$$

Therefore

$$- \big||x|- |y|\big| - |z| \le |x| - |y+z| \le \big||x| - |y|\big| + |z|$$

which implies $$\big| |x| - |y+z|\big| \le \big||x| - |y|\big| + |z|$$


Somos 05/16/2018.

Let $\; w:=-y-z,\; a:=|x|-|w|,\; b:=|x|-|y|.\;$ Using the reverse triangle inequality twice gives $\; |a| - |b| \le ||a| - |b| | \le | a-b | = | |y|-|w| | \le |y+w| = |-z| = |z|.\;$ This is $\;|a| \le |b|+|z|.\;$ Substituting for $\;a,b\;$ gives $\; ||x|-|w|| = ||x| - |y+z| | \le ||x|-|y|| + |z| \;$ which is our result.


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