Is $|f(a) - f(b)| \leqslant |g(a) - g(b)| + |h(a) - h(b)|$? when $f = \max\{{g, h}\}$ 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.
 A: 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
$$
  | f(a) - f(b) | \le |g(a) - g(b)| + |h(a) -h(b)| \, .
$$
A: 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}
