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Prove:$\forall a_1,b_1,a_2,b_2: \left|\max(a_1,b_1) - \max(a_2,b_2)\right| \le \max(\left|a_1-a_2\right|, \left| b_1-b_2 \right|) $

How can I prove it easily? Or should I just go over each case? All those cases make me confused!

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WLOG, assume $a_1\geq b_1$.

If $a_2\ge b_2$, LHS=$|a_1-a_2|$, the inequality holds.

If $a_2<b_2$, then LHS=$|a_1-b_2|$, we further separate it into two cases.

If $b_2\le a_1$ then LHS=$|a_1-b_2|\le |a_1-a_2|$, the inequality holds.

If $b_2> a_1$ then LHS=$|a_1-b_2|\le |b_1-b_2|$, the inequality holds.

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  • $\begingroup$ Thank you. You made it very clear! $\endgroup$ – AlonAlon Nov 15 '14 at 16:25
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Just some hints.

Here are some general principles that can be helpful. See if these work for you.

  1. Proving $|A| \leq B$ amounts to proving $A \leq B$ and $-A \leq B$.

  2. Proving that $\max(A,B) \leq C$ amounts to proving $A \leq C$ and $B \leq C$.

  3. Proving that $\max(A,B) \geq C$ amounts to proving $A \geq C$ or $B \geq C$.

After some manipulations and omission of symmetric cases, I believe you'll find that the inequality to be proved can be reduced to $$a_1 \leq |a_1 - a_2| + a_2.$$

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