Prove that $||a|-|b||$ is smaller or equal to $|a-b|$

Question

I am stuck with this question:

Show that $\vert \vert a\vert - \vert b\vert\vert \le \vert a-b\vert$

I had tried proving this using the following method below:

$\vert a\vert+\vert b\vert \ge \vert a-b\vert$

$-\vert a\vert-\vert b\vert \le -\vert a-b\vert$

Since $ -\vert a-b\vert = \vert b-a\vert=\vert a-b\vert$, then $-\vert a\vert-\vert b\vert \le \vert a-b\vert$

$\vert -a\vert-\vert b\vert=\vert a\vert -\vert b \vert \le \vert a-b\vert$

$\vert\vert a\vert-\vert b\vert\vert \le \vert a-b\vert$ since $\vert\vert a-b\vert\vert=\vert a-b\vert$ QED

Am I doing this correctly? Something really feels wrong here but I can't seem to find it, any help is appreciated, thank you!

Answer 1

You are wrong in this step

$$-|a|-|b| \leq |a-b| \Rightarrow |-a|-|b| \leq |a-b|$$ as $$|-a|=|a| \neq -|a|$$

---

A better way of doing this would be

$$|a|+|b| \geq |a-b|$$

Now, substitute $a=a$ and $b=a-b$ to get,

$$|a-b| \geq |b| -|a| \tag{1}$$

Also, $$|a|+|b| \geq |b-a|$$

as $$|a-b|=|-(a-b)|=|b-a|$$

Now, substitute $a=b-a$ and $b=b$ to get,

$$|b-a| \geq |a|-|b|\tag{2}$$

Now, as $$|b-a|=|a-b|$$

We may write $(2)$ as

$$|a-b| \geq |a|-|b|\tag{3}$$

because

$$|a-b|=|-(a-b)|=|b-a|$$ From $(1)$ and $(3)$, we can conclude that

$$|a-b| \geq ||a|-|b||$$

as $$||a|-|b||= \begin{cases} |a|-|b|, \text{ if }|a| \geq |b|\\ |b|-|a|, \text{ if }|b| \geq |a|\\ \end{cases}$$

Answer 2

The "basic idea" is that $||a| - |b||$ is a positive difference of terms and $|a - b|$ is either an equal positive difference of terms (if $a$ and $b$ are the same sign), or (if $a$ and $b$ are opposite signs) is a positive sum of terms which (neither are zero) is bigger than the difference.

====

Are you familiar with "wolog" = "without loss of generality"?

Since $|a - b| = |b - a|$ (and $||a| - |b|| = ||b| - |a||)$, we can-- if we need to-- switch the variable $a$ with $b$ and $b$ with $a$ and the result will be the same.

As either $|a| \ge |b|$ or $|b| \ge |a|$. If $|b| \ge |a|$ we can switch and rename the variable to get $|a| \ge |b|$.

So we say: "without loss of generality (or wolog for short), we may assume $|a| \ge |b|$".

Likewise as $|-x| = |x|$ (and $||-a| - |-b|| = ||a|-|b||$, $|a - b| = |-(a-b)| = |(-a) - (-b)|$ so

without loss of generality, we may assume $a \ge 0$ (because if $a < 0$ we can just replace $a$ and $b$ with $-a$ and $-b$.)

So wolog we may assume $|a| = a \ge |b| \ge 0$.

So $|a| - |b| \ge 0$. So $||a| - |b|| = |a| - |b| = a - |b| \ge 0$.

So the entire statement boils down to $||a|-|b||= a - |b| \le a + |b|$ which is obvious.

$|a - b| = a - b = a -|b|$ if $b \ge 0$, and $|a - b| = a - b = a + |b| > a - |b| if $b < 0$.

So $|a|-|b|| = a - |b| \le a - b = |a-b|$ with equality holding if and only if $b \ge 0$.

QED

But if that "wolog" seems a bit ... much...

There's nothing wrong with breaking it into cases:

Case 1: $a \ge 0$ and $|a| \ge |b|$.

see above.

Case 2: $a < 0$ and $|a| \ge |b|$.

Let $a' = -a$ and $b' = -b$, and do Case 1.

Case 3: $|b| < |a|$.

Let $a' = b$ and $b'=a$. Then do either case 1 or 2.

Answer 3

Although you probably intended $a,b$ to be real numbers, we can actually prove this result much more generally; $a,b$ can be points in any vector space. Then $\left|a\right|,\left|b\right|,\left|a-b\right|$ are the side lengths of the triangle whose vertices are the origin and $a$ and $b$. If the first two of these side lengths differed by more than the third side length, the larger of the first two side lengths would be greater than the sum of the other two side lengths, which contradicts the triangle inequality. So the crux of the problem is verifying the triangle inequality for the norm $\left|\cdot\right|$. (In particular, it needn't be the Euclidean $L^2$ norm.) See also here.

Answer 4

WLOG, $a\ge0$ (because you can change the signs of $a$ and $b$ simultaneously).

Then

• if $b\ge 0$, $||a|-|b||=|a-b|\le|a-b|$ (trivially),
• if $(-b)>0$, $||a|-|b||=|a-(-b)|\le|a+(-b)|$, because the difference of two positive cannot exceed their sum.