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If $|X| \leq K$ and $|Y| \leq K$, then how can I show that $|X-Y| \leq 2K$? It seems very obvious intuitively but I am not able to use the triangle inequality here. Does anyone have any ideas? thanks!

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  • $\begingroup$ Why cann't you use the triangle inequality? $\endgroup$ – Hetebrij Nov 28 '15 at 21:24
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    $\begingroup$ @Hetebrij Yes he can! $\endgroup$ – user60589 Nov 28 '15 at 21:25
  • $\begingroup$ If you cannot (are not allowed to) use the triangle inequality, you will in effect have to prove the triangle inequality. $\endgroup$ – André Nicolas Nov 28 '15 at 21:27
  • $\begingroup$ @user60589 No, we can, but I want to understand why user136503 thinks he is not able to use the triangle inequality. Or if it is forbidden by some power. $\endgroup$ – Hetebrij Nov 28 '15 at 21:29
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    $\begingroup$ try to use the triangle inequality with $|X+(-Y)|$ $\endgroup$ – Mirko Nov 28 '15 at 21:37
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First observe that $|x-y| \leq |x| + |y|$ since $|x-y| = |x + (-y)| \leq |x| + |(-y)| = |x| + |y|$ according to the triangle inequality.

Therefore, if we suppose that for arbitrary real numbers $x, y,$ and $k$ the inequalities $|x| \leq k$ and $|y| \leq k$ are true, then

$$|x-y| \leq |x| + |y| \leq 2\cdot k \implies |x-y| \leq 2 \cdot k$$

as desired.

$\blacksquare$

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$$|X - Y| \le |X| + |Y| \le K + K = 2K$$

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  • $\begingroup$ Reproducing (after faulty versions) @Benedict's answer. $\endgroup$ – Did Nov 29 '15 at 8:16
  • $\begingroup$ If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review $\endgroup$ – Ian Miller Nov 29 '15 at 8:52
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    $\begingroup$ @IanMiller I don't have a new question. Me saying 'I think' was because I found this a little too easy. It seemed like a simple application of the triangle inequality. I found the question kind of strange consider user136503 has posted a lot of advanced probability problems $\endgroup$ – BCLC Nov 29 '15 at 8:55

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