How to prove $||A| - |B|| \le |A -B|$ Suppose A and B are real numbers, prove following inequality:
$$
||A| - |B|| \leq |A - B|
$$
How to prove this inequality？ Thanks.
 A: \begin{align}
|P+Q| & \le |P|+|Q| \\[8pt]
\text{Therefore }|(A-B)+B| & \le |A-B| + |B|. \\[8pt]
|A| & \le |A-B|+|B| \\[8pt]
|A| - |B| & \le | A - B|
\end{align}
By the same method, show that $|B|-|A|\le|A-B|$ and then you've got it.
A: Note that $|x-y|$ is the distance between the points $x$ and $y$ on the real number line. We have to prove that $|A|$ and $|B|$ are at least as close as $A$ and $B$. If $A$ and $B$ are of the same sign, $|A|$ and $|B|$ are as close as $A$ and $B$. If they are of different signs, $|A|$ and $|B|$ are closer than $A$ and $B$ since the distance between $|A|$ and $|B|$ now equals the distance between $A$ and $-B$ (assuming $|A|>|B|$) where $-B$ lies between $A$ and $B$ and therefore is closer to $A$ than $B$ is.
A: This is called the inequality of the uniform continuty of $|\cdot|$ in a more general normed linear space than $\mathbb{R}$ any norm satisfies that condition. 
For alternative proof, squaring both sides to obtain and equivalent inequality 
$$-2|A||B|\leq 2AB$$ which is trivially true.
But i like more the Hardy's answer :) 
