Why is absolute difference of max of functions less than max of absolute difference of functions

We have that

$$f(x)\le |f(x)-g(x)|+g(x).$$ Thus

$$\max f(x)\le \max (|f(x)-g(x)|+g(x))\le \max |f(x)-g(x)| +\max g(x).$$ So

$$\max f(x)-\max g(x)\le \max |f(x)-g(x)|.$$ Changing the roles of $f$ and $g$ we have that

$$\max g(x)-\max f(x)\le \max |f(x)-g(x)|.$$ Thus we are done.

Alternatively (to the other answer), to remember which way this inequality goes, compare $\sin(x)$ and $-\sin(x)$ (on, say , $(-\infty, \infty)$, although many non-tricky choices are equivalent). The maximum of both functions is $1$, so the difference of the maxima is $0$. The maxima of one coincides with the minima of the other and vice versa, so $|\sin(\pi/2) - (-\sin(\pi/2))| = 2$ is the maximum of the absolute differences.

The inobviousness of this sort of result seems to be thinking that the maxima of both functions happen at the same places in the domain.