Q: $f(n)$ and $g(n)$ are asymptotically non-negative functions. prove that $\max(f(n), g(n)) = \Theta(f(n)+g(n))$
my proof as follows:
If $\exists n_0$ such that $f(n)\geq g(n)\geq 0$ $\forall n \geq n_0$, then $$0\leq\frac{1}{2}(f(n)+g(n))\leq f(n)=\max(f(n),g(n)) \leq 1(f(n)+g(n))$$
If $\exists n_0$ such that $g(n)> f(n) \geq 0$ $\forall n \geq n_0$, then $$0\leq\frac{1}{2}(f(n)+g(n))\leq g(n)=\max(f(n),g(n)) \leq 1(f(n)+g(n))$$
Thus, by $\Theta$ definition, $\max(f(n), g(n)) = \Theta(f(n)+g(n))$
I am not sure whether my proof is rigorously enough and logically correct.
Can I suppose or assume "if $\exists n_0$ ..." in the above-mentioned proof?