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This question already has an answer here:

I guess, I know how to solve inequalities with absolute value, but I have problems with this one.

$||a|-|b||\le |a-b|$

$a,b\in \mathbb{R}$

I tried to solve the inequation like this:

case1: $a>0$

case2: $a<0$

I started with case 1. than we have two possibilities $b<0$ and $b>0$

Firstly I took $b<0$ and I had another two possibilities

$|a+b|>0$

$|a+b|<0$

I take $|a+b|>0$ and in this case I had two possibilites according to to right side of the inequation

$|a-b|>0 ........... a+b\le a-b ......b\le-b$

$|a-b|<0..........a+b\le -a+b .......a\le -a$

I can do the same with other possibilites, but how do I know, if my solution (or even method) is right? there are so many conditions, that I am lost in them.

Thank you for your time.

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marked as duplicate by Martin Sleziak, Community Oct 10 '16 at 6:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$ ||a|-|b||\le |a-b| $$ iff $$ ||a|-|b||^2\le |a-b|^2 $$ iff $$ a^2 -2|ab| +b^2 \leq a^2-2ab+b^2 $$ iff $$ ab \leq |ab|, $$ and the last inequality is always true.

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    $\begingroup$ Thanks. Just one think is a bit unclear to me. When going from 2nd to the 3rd inequation, how do you know, that when you square absolute value, the roots won`t cange ? is it somehow connected with positivity of the expression in the absolute value? $\endgroup$ – martina Oct 9 '16 at 12:18
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    $\begingroup$ @martina: Yes. An absolute value is non-negative, and if $x,y$ are non-negative numbers, then $x \leq y$ iff $x^2 \leq y^2$. $\endgroup$ – Evan Aad Oct 9 '16 at 12:26
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There is a simpler solution.

First, notice that the inequality you want to show is equivalent to $$|a| - |b|\leq |a-b| \text{ and } |b| - |a|\leq |a-b|$$ Let's see how to show the first one. We have $|a| = |(a-b) + b|$. Using the triangular ineaquality, this implies that $|a|\leq |a-b| + |b|$, thus $|a|-|b|\leq |a-b|$. You can show the other inequality exactly the same way.

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    $\begingroup$ Thanks. Can you add some more comments to me understand your solution more clearly? I am lost there, when you used triangular inequality...... $\endgroup$ – martina Oct 9 '16 at 12:31
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    $\begingroup$ $|x+y|\leq |x| + |y|$. I just applied this to $x=a-b$ and $y=b$. $\endgroup$ – Augustin Oct 9 '16 at 13:16
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Instead of considering signs of $a$ and $b$ separately, you can consider their relative signs: if they have the same sign (or one is zero), then the two sides are equal, while if they have different (nonzero) signs, the right hand side is bigger (equal to sum of the absolute values).

If this is not obvious to you, then you can simplify a bit by assuming that $a$ is nonnegative: you can do that because flipping signs of $a$ and $b$ both does not change either side.

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