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.
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2$\begingroup$ It follows from the triangle inequality, letting $y = -z$ you get that $$|x + y| = |x - z| \leq |x| + |-z| = |x| + |z|$$ $\endgroup$– Edward EvansSep 7, 2016 at 14:57
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1$\begingroup$ can you tell us what you've tried? Why do you think your proof is not correct? $\endgroup$– LudolilaSep 7, 2016 at 14:57
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2$\begingroup$ This with $-y$ instead of $y$, because $\lvert -y\rvert=\lvert y\rvert$. $\endgroup$– user228113Sep 7, 2016 at 14:59
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1$\begingroup$ Or this. $\endgroup$– user228113Sep 7, 2016 at 15:01
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$\begingroup$ You don't bother to search for the question. And you are asking similar questions without showing any work or sharing your thoughts. $\endgroup$– StubbornAtomSep 7, 2016 at 15:06
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3 Answers
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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 $$ |x-y|\leq |x|+|y| $$
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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 $|-Y|=|Y|$ , $$|X-Y| \leq |X|+|Y|$$
answered Sep 7, 2016 at 15:14
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$$\sqrt{(x-y)^2}\leqslant \sqrt{x^2}+\sqrt{y^2}$$ $$\sqrt{x^2-2xy+y^2}\leqslant \sqrt{x^2}+\sqrt{y^2}$$ $$x^2-2xy+y^2\leqslant x^2+y^2+2\sqrt{y^2x^2}$$
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$\begingroup$ 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). $\endgroup$– TravisJSep 7, 2016 at 15:23