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.

It follows from the triangle inequality, letting $y = -z$ you get that
$$|x + y| = |x - z| \leq |x| + |-z| = |x| + |z|$$ — Edward Evans, Sep 07 '16 at 14:57
can you tell us what you've tried? Why do you think your proof is not correct? — Ludolila, Sep 07 '16 at 14:57
This with $-y$ instead of $y$, because $\lvert -y\rvert=\lvert y\rvert$. — , Sep 07 '16 at 14:59
You don't bother to search for the question. And you are asking similar questions without showing any work or sharing your thoughts. — StubbornAtom, Sep 07 '16 at 15:06

score 5 · Accepted Answer · answered Sep 07 '16 at 14:57

5

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

answered Sep 07 '16 at 14:57

Frank

2,348

score 1 · Answer 2 · answered Sep 07 '16 at 15:14

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

E.H.E · Answer 3 · 2016-09-07T15:30:32.797

-1

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

edited Sep 07 '16 at 15:30

answered Sep 07 '16 at 15:01

E.H.E

23,280

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). – TravisJ Sep 07 '16 at 15:23

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

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