Prove: $||x|-|y||\leq |x+y|$ I have to prove: $||x|-|y||\leq |x+y|\leq |x|+|y|$ I have already proved $|x+y|\leq |x|+|y|$.
I saw proofs for $||x|-|y||\leq |x-y|$, can I use the proof for  $||x|-|y||\leq |x-y|$ and just add that $|x-y|\leq|x+y|$?
 A: 
I have already proved $|x+y|\leq |x|+|y|$.

One may observe that
$$
|x|=|x-y+y|\le |x-y|+|y|
$$ swapping $x$ and $y$ gives
$$
|y|=|y-x+x|\le |y-x|+|x|,
$$ from which one gets
$$
||x|-|y||\leq |x-y|
$$ then making $y \to-y$ we obtain
$$
||x|-|y||\leq |x+y|.
$$
A: Hint: $$||x|-|y||\le |x+y| \Leftrightarrow (||x|-|y||)^2\le (|x+y|)^2$$
A: $$\left| x \right| =\left| x+y-y \right| \le \left| x+y \right| +\left| y \right| \quad \Rightarrow \left| x \right| -\left| y \right| \le \left| x+y \right| \\ \left| y \right| =\left| y+x-x \right| \le \left| y+x \right| +\left| x \right| \quad \Rightarrow \left| y \right| -\left| x \right| \le \left| x+y \right| $$
combine all together then $$||x|-|y||\leq |x+y|\\ \\ $$
A: If you already know how to show that $\left||u|-|v|\right|\leq |u-v|$, then let $u=x,v=-y$.
A: $$(\lvert\lvert x\rvert-\lvert y \rvert \rvert)^2 \le (\lvert x + y \rvert)^2$$
$$\lvert x\rvert^2 + \lvert y\rvert^2 - \lvert 2xy\rvert\le x^2+y^2+2xy$$
$$-\lvert xy\rvert \le  xy$$
Which is of course true by the definition of the absolute value
