Prove those inequalities are true

Question

I want to prove that those inequalities are true for $a, b ∈ R$: $$ |a + b| ≤ |a| + |b| $$ $$ ||a| − |b|| ≤ |a − b| $$ $$ |a − b| ≤ |a − c| + |c − b| $$

Now I can see that they are true, and I could argue it with words, but how would you prove it on paper using mathematic notation?

This very much depends on what you are already given. Could you provide us with some background on what you already know and what you tried so far? — Jendrik Stelzner, Jan 09 '16 at 15:28
Think about what the absolute value of something means. It may help to consider that $|x|=\max{x,-x}$. — J.Cork, Jan 09 '16 at 15:32
@JendrikStelzner what I would try is take every case possible - a positive, b positive | a negative, b negative | a positive, b negative | a negative, b positive. Now I would show what it would look like after getting rid of the abs() and show that the statement is true for all cases. — Mykybo, Jan 09 '16 at 15:34
Also, the second two can be proven using the first (see if you can do this), so then all you need to do is prove the first! — J.Cork, Jan 09 '16 at 15:34
@J.Cork though I am not sure how I would use the first equation to prove the other two? — Mykybo, Jan 09 '16 at 15:38
The idea of case distinction is right, but you should choose your cases so that in each case you can evaluate the term $|x-y|$ using whatever definition of $|\cdot|$ you’re using. And as J.Cork already pointed out the last two equations follow from the first by some playing around. — Jendrik Stelzner, Jan 09 '16 at 15:40
@Mykybo In both cases, try applying the first inequality to things such as "$|a|=|a-b+b|$" etc — J.Cork, Jan 09 '16 at 15:44

Simply Beautiful Art · Answer 1 · 2016-01-09T16:41:22.030

Here's a hint:

$$|x|=x$$$$x\ge0$$$$|y|=-y$$$$y<0$$

Try using that. Treat every case for $x$, the stuff inside the absolute values, individually until you find the answer.

An idea, noted in the comments by N. F. Taussig, is as follows:

$$|a|=+\sqrt{a^2}$$

So we have the following (easiest to use an equal sign):

$$\sqrt{(a+b)^2}=|a+b|=\sqrt{a^2}+\sqrt{b^2}$$

$$(a+b)^2=(\sqrt{a^2}+\sqrt{b^2})^2$$

$$a^2+2ab+b^2=a^2+2\sqrt{a^2b^2}+b^2$$

$$2ab=2\sqrt{a^2b^2}$$

We have the following scenarios, $a,b\ge0$, then we have:

$$2ab=2ab$$

If we have $a,b<0$, then we have:

$$2(-a)(-b)=2ab$$

If we have $a$ or $b$ $<0$, then we have:

$$2(-a)b=2a(-b)<0<2\sqrt{a^2(-b)^2}=2\sqrt{(-a)^2b^2}=2\sqrt{a^2b^2}=2ab$$

Using the $<0<$ in the middle helps define what is happening.

The same would have happened if had used the definition $|x|=x,x\ge0$ and $|x|=-x,x<0$.

Essentially, instead of proving it with words, we break it in to different scenarios describing when the values will be positive or negative by simply using $x\ge0$ or $x<0$.

This problem was posed for real numbers, however it does still hold for complex numbers. In fact for any metric-vector-space, where $|x|=d(x,0)$, these inequalities must hold. — J.Cork, Jan 09 '16 at 15:52

Prove those inequalities are true

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