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I have to show that for $a,b,c,d \in \mathbb{R}$ $$|a\vee b - c\vee d| \leq |a-c|\vee |b-d|$$ I know this can be showed using cases, but I need help with a proof that doesn't involve cases. I found out this inequality can be written as: $$|a\vee b - c\vee d| \leq (a\vee c - a \wedge c) \vee (b \vee d - b \wedge d)$$ Since $a\vee c= \frac{(a+c) + |a-c|}{2}$ and $a\wedge c= \frac{(a+c) - |a-c|}{2}$

I guess the proof uses this form of the inequality, but I don't know how to write such proof.

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This is basically just repeated applications of the triangle inequality; namely, we have $$ \lvert \max(a,b) - \max(c,d) \rvert = \bigg\lvert \frac{(a-c) + (b-d) + \lvert a - b \rvert - \lvert c - d \rvert}{2}\bigg\rvert \leq \frac{(\lvert a - c \rvert + \lvert b - d \rvert) + \lvert \lvert a-b \rvert - \lvert c - d \rvert\rvert}{2}.$$ But, $$\lvert \lvert a-b \rvert - \lvert c - d \rvert \rvert \leq \lvert \lvert a-c \rvert + \lvert b-c \rvert - (\lvert c-b \rvert + \lvert b-d \rvert) \rvert = \lvert \lvert a-c \rvert - \lvert b - d \rvert \rvert.$$ That should give the desired result.

Edit: For the last step, I used $$\lvert a - b \rvert = \lvert a - c + c - b \rvert \leq \lvert a-c \rvert + \lvert b - c \rvert.$$ You use that same trick on both terms in the absolute value.

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  • $\begingroup$ What form of the triangle inequality you used at that last step? $\endgroup$ – MMM Mar 15 at 18:22
  • $\begingroup$ I added an edit to (hopefully) address your question. $\endgroup$ – Gary Moon Mar 15 at 18:26

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