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Let $f = \max\{{g, h}\}$ where all 3 of these functions map $\mathbb{R}$ into itself. Is it true that $|f(a) - f(b)| \leqslant |g(a) - g(b)| + |h(a) - h(b)|$? I'm thinking it can be proven by cleverly adding and subtracting inside of the absolute value and then using the triangle inequality, but i'm completely stuck.

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If $g(a) \ge h(a)$ then (since $f(b) \ge g(b)$): $$ f(a) - f(b) = g(a) - f(b) \le g(a) - g(b) \le | g(a) - g(b) | \le \max \{ |g(a) - g(b)|, |h(a) -h(b)| \} \, . $$

The same is true if $g(a) \le h(a)$, and by symmetry (exchange $a$ with $b$) it follows that $$ | f(a) - f(b) | \le \max \{ |g(a) - g(b)|, |h(a) -h(b)| \} $$ for all $a, b$. This is a stronger inequality than the desired $$ | f(a) - f(b) | \le |g(a) - g(b)| + |h(a) -h(b)| \, . $$

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  • $\begingroup$ Perfect, sir. This is a tricky problem, therefore +1 for your efforts. I spent 20 minutes trying to answer this, and the answer just kept fading away from me. $\endgroup$ – астон вілла олоф мэллбэрг Mar 13 '16 at 9:49
  • $\begingroup$ @abdefghijklmnopqrtxyz-stoo: Thank you! Actually it also took me a while, from "this must be false" to "this must be true" over a wrong proof until I got this. $\endgroup$ – Martin R Mar 13 '16 at 9:53
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If you dont want to deal with multiples cases :

\begin{array}{lcl} |f(a)-f(b)| & = & \left| \max\{g(a),h(a)\} - \max\{g(b),h(b)\} \right| \\ & = &\left| \frac{g(a)+h(a)+|g(a)-h(a)|}{2} - \frac{g(b)+h(b)+|g(b)-h(b)|}{2} \right| \\ & = & \left| \frac{g(a)-g(b)}{2} + \frac{h(a)-h(b)}{2} + \frac{|g(a)-h(a)|-|g(b)-h(b)|}{2} \right| \\ & \le & \frac{1}{2}|g(a)-g(b)| + \frac{1}{2}|h(a)-h(b)| + \frac{1}{2}\Bigl\lvert |g(a)-h(a)|-|g(b)-h(b)|\Bigl\lvert \\ & \le & \frac{|g(a)-g(b)|}{2} + \frac{|h(a)-h(b)|}{2} + \frac{1}{2} \Bigl\lvert(g(a)-h(a))-(g(b)-h(b)) \Bigl\lvert\\ & \le & \frac{|g(a)-g(b)|}{2} + \frac{|h(a)-h(b)|}{2} + \frac{1}{2} \Bigl\lvert (g(a)-g(b))-(h(a)-h(b))\Bigl\lvert \\ & \le & \frac{|g(a)-g(b)|}{2} + \frac{|h(a)-h(b)|}{2} + \frac{1}{2}\left(|g(a)-g(b)|+ |h(a)-h(b)|\right) \\ & \le & |g(a)-g(b)|+ |h(a)-h(b)| \\ \end{array}

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