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Is there a clever way to show that

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

I believe I can think of a way to solve this using cases, but the book I'm working out of said that "A very short proof is possible if you write things in the right way" which has motivated me to figure out this quick way, but I've failed thus far and any help would be appreciated

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    $\begingroup$ Hint: Write $x=x-y+y$ on the LHS. $\endgroup$ – Git Gud Jul 24 '14 at 17:23
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$\left|x\right|=\left|x-y+y\right|\leq\left|x-y\right|+\left|y\right|$ so $\left|x\right|-\left|y\right|\leq\left|x-y\right|$

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    $\begingroup$ that is pretty clever. thanks! $\endgroup$ – Math_Illiterate Jul 24 '14 at 17:39
  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Jul 24 '14 at 17:51
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Your inequality is equivalent to the triangle inequality, so using it proves your inequality in one step. (See drhab's answer.)

Without using the triangle inequality, we can still find a proof that doesn't need cases:

Your inequality is equivalent to $$|x| \le |x-y| + |y|.$$ Since both sides are non-negative, this is equivalent to $$|x|^2 \le (|x-y| + |y|)^2.$$ Calculate the squares and get rid of $x^2$ on both sides to get $$0\le 2y^2-2xy + + 2|(x-y)y|,$$ which is equivalent to $$2(x-y)y \le 2|(x-y)y|,$$ which is true since $z\le|z|$ for all $z\in\mathbb R$.

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