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 $|x-y|\leq|x+y|$?
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No, because $|x - y| \le |x + y|$ is wrong. Apply the inequality you know with $-y$ replacing $y$. – Jul 05 '16 at 18:35
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1It's not true that $|x-y|\leq |x+y|$. Try $x=1,y=-1$. – Thomas Andrews Jul 05 '16 at 18:44
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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 $$ ||x|-|y||\leq |x+y|. $$
Olivier Oloa
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With this I can get the proof:$ ||x|-|y||\leq |x-y|$ but how can I connect it to $|x+y|$? – gbox Jul 05 '16 at 18:39
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1@Roby5 Two steps: $||x|-|y||\le |x-y|$. Then set $y=-b$ giving $||x|-|b||\le |x+b|$ – Olivier Oloa Jul 05 '16 at 18:56
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$$\left| x \right| =\left| x+y-y \right| \le \left| x+y \right| +\left| y \right| \quad \Rightarrow \left| x \right| -\left| y \right| \le \left| x+y \right| \\ \left| y \right| =\left| y+x-x \right| \le \left| y+x \right| +\left| x \right| \quad \Rightarrow \left| y \right| -\left| x \right| \le \left| x+y \right| $$ combine all together then $$||x|-|y||\leq |x+y|\\ \\ $$
haqnatural
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If you already know how to show that $\left||u|-|v|\right|\leq |u-v|$, then let $u=x,v=-y$.
Thomas Andrews
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$$(\lvert\lvert x\rvert-\lvert y \rvert \rvert)^2 \le (\lvert x + y \rvert)^2$$
$$\lvert x\rvert^2 + \lvert y\rvert^2 - \lvert 2xy\rvert\le x^2+y^2+2xy$$
$$-\lvert xy\rvert \le xy$$
Which is of course true by the definition of the absolute value
