# 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 $g(a) \ge h(a)$ 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 $$| f(a) - f(b) | \le |g(a) - g(b)| + |h(a) -h(b)| \, .$$