Showing that $\max\{f+g\} \leq \max f + \max g$
Solution 1:
For all $x$, $$f(x)\le \max f(x)$$ and $$g(x)\le\max g(x).$$ Now add the two together.
Solution 2:
let $\max g(z):=g(z^*)$, $\max f(z):=f(z^*)$ then $$f(z)\le f(z^*)$$ $$g(z)\le g(z^*) $$ then $$\forall z : (f+g)(z)=f(z)+g(z)\le f(z^*)+g(z^*)$$ $$\max(f+g)(z)\le f(z^*)+g(z^*)=\max f(z)+\max g(z)$$