# Prove: $|x-y|\leq |x|+|y|$ [duplicate]

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.

• It follows from the triangle inequality, letting $y = -z$ you get that $$|x + y| = |x - z| \leq |x| + |-z| = |x| + |z|$$ Sep 7, 2016 at 14:57
• can you tell us what you've tried? Why do you think your proof is not correct? Sep 7, 2016 at 14:57
• This with $-y$ instead of $y$, because $\lvert -y\rvert=\lvert y\rvert$.
– user228113
Sep 7, 2016 at 14:59
• Or this.
– user228113
Sep 7, 2016 at 15:01
• You don't bother to search for the question. And you are asking similar questions without showing any work or sharing your thoughts. Sep 7, 2016 at 15:06

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

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

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

• It appears to me that the OP is asking for a proof of the triangle inequality, not an "assertion" that it holds. Also, your statements are false (when $x=y=0$ for example). Sep 7, 2016 at 15:23