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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!

Bérénice
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Derp
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    $\vert \vert a \vert - \vert b \vert \vert \leq \vert a - b \vert$ is equivalent to $- \vert a - b \vert \leq \vert a \vert - \vert b \vert \leq \vert a - b \vert$. Show the latter. – Steven Harding Jun 02 '16 at 15:43
  • Hold on, so are we using the fact that $-\vert a\vert \le a \le \vert a\vert$ to show the above equation? – Derp Jun 02 '16 at 15:46
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    −|a−b|=|b−a| is false. – Doug M Jun 02 '16 at 15:47
  • We are using the fact that $\vert a \vert \leq b$ if and only if $-b \leq a \leq b$. – Steven Harding Jun 02 '16 at 15:47
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    Two ways you can go by (more than 2, I will offer 2). First go through all of the cases (aka brute force) a>b>0, b>a>0, a>0>b, etc. option 2, square both sides. – Doug M Jun 02 '16 at 15:50
  • If I use the brute force method would that mean I need to show 6 possibilities? Or just 5? Since 0>a>b is similar to 0>b>a. – Derp Jun 02 '16 at 15:59
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    If you do brute force, you can do exactly what you have done above. Acknowledge the case and explain in one line that the algebra will be so similar to a previous case that you won't step through the details a second time. – Doug M Jun 02 '16 at 16:08
  • Understood, then about the squaring both sides method, how do I start? Do I square $\vert a\vert +\vert b\vert \ge \vert a-b\vert$? – Derp Jun 02 '16 at 16:10
  • "Since −|a−b|=|b−a|=|a−b|, then −|a|−|b|≤|a−b|" This isn't true $-|a - b| \ne |b-a| = |a-b|$ and obviously $-|a-b| \ne |a - b|$. Instead -|a-b| = -|b-a|. The result -|a| - |b| <= |a - b| is trivial as the LHS is less is negative or zero and the RHS is positive of zero. – fleablood Jun 02 '16 at 16:19
  • "||a|−|b||≤|a−b| since ||a−b||=|a−b| QED" ?????? I don't see how ||a-b|| = |a-b| implies ||a|-|b|| ≤|a-b| and I don't see how any of the stuff you did above goes from one statement to the next. Your proof seems to be unconnected statements; the first two true and the only two connected in any way, the third false with a trivially true conclusion that doesn't follow, The fourth is true. The fifth the conclusion from no good reason. – fleablood Jun 02 '16 at 16:26

4 Answers4

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

Roby5
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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.
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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.

fleablood
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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.

J.G.
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