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.

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}
