Prove those inequalities are true 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?
 A: 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 $x<0$.
