Prove: $|x-y|\leq |x|+|y|$

Question

Can you help me to prove that $$|x-y|\leq |x|+|y|$$ I get a proff of this equality, but it's very short and I don't know if it's correct.

Answer 1

Take the square root of this: $$ (x-y)^2=x^2+y^2-2xy\leq x^2+y^2+2|x||y|=(|x|+|y|)^2 $$ and get $$ |x-y|\leq |x|+|y| $$

Answer 2

$$\sqrt{(x-y)^2}\leqslant \sqrt{x^2}+\sqrt{y^2}$$ $$\sqrt{x^2-2xy+y^2}\leqslant \sqrt{x^2}+\sqrt{y^2}$$ $$x^2-2xy+y^2\leqslant x^2+y^2+2\sqrt{y^2x^2}$$

Answer 3

Clearly $|x| = \max\{x,-x\}$

Thus $\pm x ≤ |x|$.

Then you can observe that :

\begin{align*} x + y &≤ |x| + y ≤ |x| + |y|,\quad\text{and}\\ -x - y &≤ |x| -y ≤ |x| + |y|. \end{align*}

So we have that $|x+y| \leq |x|+|y|$

Now put $x=X$ and $y=-Y$ ,

Thus $$|X-Y| \leq |X|+|-Y|$$

Since $|-Y|=|Y|$ , $$|X-Y| \leq |X|+|Y|$$