Formally prove that $\Theta(\max(f,g)) = \Theta(f+g)$ I am having a hard time proving that $\Theta(\max(f,g)) = \Theta(f+g) $
where
$(f+g)(n) = f(n) + g(n) $
and
$(\max{f,g})(n) = \max(f(n), g(n))$
I know that $\Theta$ is the combination of the upper and lower bounds, but I can't seem to prove this. It's hard for me to see how $\Theta$ of the max of two functions can be equivalent to $\Theta$ of the two functions added together.  Any guidance would be appreciated. Let me know if I can provide more info.
This question is similar, but didn't quite help.
 A: Certainly, $\max(f,g) \leq f+g$ so $\max(f,g) = O(f+g)$, and it only remains to establish the lower bound.
If $f = \Theta(g)$, the statement is trivial, so assume $f = O(g)$ and then $2 \max(f,g) = 2g \geq f+g$, which implies $\max(f,g) = \Omega(f+g)$, as desired.
A: If $f,g \geq 0$, then $\frac{1}{2}(f+g) \leq \max(f,g) \leq (f+g)$. Rearranging gives $ \max(f,g) \leq (f+g) \leq  2 \max(f,g)$.
It follows that $\Theta(f+g) = \Theta(\max(f,g))$.
Addendum: When we write $a \in \Theta(b)$, it means that there exists $\underline{k}, \overline{k} >0$, and $N$ such that if $n\geq N$, then $\underline{k} b(n) \leq a(n) \leq \overline{k} b(n)$. Note that if $a \in \Theta(b)$, then $\frac{1}{\overline{k}} a(n) \leq b(n) \leq \frac{1}{\underline{k}} a(n)$, and so $b \in \Theta(a)$, hence it is an equivalence relation.
The above shows that if we choose $\underline{k} = \frac{1}{2}$, $\overline{k} = 1, N=1$, then $\underline{k}(f(n)+g(n)) \leq \max(f(n),g(n)) \leq \overline{k}(f(n)+g(n))$, and hence $n \mapsto \max(f(n),g(n)) \in \Theta(n \mapsto f(n)+g(n))$, or, more colloquially, 
$\max(f,g) \in \Theta(f+g)$, and by the above remark, we also have $f+g \in \Theta(\max(f,g))$. 
A: Just hope this clarifies a bit more:
You want to show that $h(x)=\max(f(x),g(x))=\Theta(f(x)+g(x))$. Observe two cases:
Case1: 
$h(x)=f(x)$, hence $f(x) \geq g(x)$. Then $2 h(x) = 2 f(x) \geq f(x) + g(x)$,
$\Leftrightarrow h(x) \geq \frac{f(x)+g(x)}{2}$
$\Leftrightarrow h(x) = \Omega(f(x)+g(x))$
This gives the lower bound. 
Next, $h(x)=f(x) \leq f(x)+g(x)=O(f(x)+g(x))$
Hence, $h(x)=\Theta(f(x)+g(x))$
Case 2:
$h(x)=g(x)$. Similar to Case 1. 
