Prove that $\mathcal{O}(f_{1}(x)+ \dots +f_{n}(x))= \mathcal{O}(\max(f_{1}(x), \dots ,f_{n}(x)))$ I want to prove the following that based on maximum rule of functions:
$$\mathcal{O}(f_{1}(x)+ \dots  +f_{n}(x))= \mathcal{O}(\max(f_{1}(x), \dots  ,f_{n}(x)))$$
the base prove is for each 2 functions :
$$\mathcal{O}(f(n)+g(n))= \mathcal{O}(\max(f(n),g(n)))$$
Note that the functions are not negative.

I would like to get some advice how to prove for a set of functions with induction.
what I tried to do is to take each 2 functions, get the max between them and check the result with the next function, there is another way to think about that?
thanks!

Another way would be to go back to the definition to prove $g \in \mathcal{O}(f_{1}(x)+ \dots  +f_{n}(x)) \leftrightarrow g \in \mathcal{O}(\max(f_{1}(x), \dots  ,f_{n}(x))$:
$g \in \mathcal{O}(f_{1}(x)+ \dots  +f_{n}(x))$ implies:
$\exists k>0 \; \exists n_0 \; \forall n>n_0 \; g(n) \leq (f_{1}(x)+ \dots  +f_{n}(x))\cdot k$ implies:
$\exists k>0 \; \exists n_0 \; \forall n>n_0 \; g(n) \leq (\max(f_{1}(x), \dots  ,f_{n}(x))\cdot kn$ implies:
$\exists K>0 \; \exists n_0 \; \forall n>n_0 \; g(n) \leq (\max(f_{1}(x), \dots  ,f_{n}(x))\cdot K$ implies:
$g \in \mathcal{O}(\max(f_{1}(x), \dots  ,f_{n}(x))$
Since $\max(f_{1}(x), \dots  ,f_{n}(x)) \leq f_{1}(x)+ \dots  +f_{n}(x)$, the reciprocal is straightforward.
