# how can be prove that $\max(f(n),g(n)) = \Theta(f(n)+g(n))$

how can be prove that $\max(f(n),g(n)) = \Theta(f(n)+g(n))$

though the big O case is simple since $\max(f(n),g(n)) \leq f(n)+g(n)$

edit : where $f(n)$ and $g(n)$ are asymptotically nonnegative functions.

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It is false by the common mathematical definition: $f(n) = n, g(n) = -n$. What definition of BigOh are you using? (I know you have tagged it algorithms, so I believe I can guess, but just making sure...) – Aryabhata Dec 22 '10 at 1:49
@Jonas Yes.they are non negative. – Bunny Rabbit Dec 22 '10 at 1:53

Hint...: $2 + 100 \le 100 + 100$
Use the fact that: $\max(f,g) \le f + g \le 2\max(f,g)$ when $f,g$ are non-negative
i don't know whatever the constant n0 we choose , the $f(n0)+g(n0) < = max(f(n0),g(n0))$ is not going to hold, since they are non negative. – Bunny Rabbit Dec 22 '10 at 2:13
only if the above condition holds for some constant n0 then we can prove the $\omega$ case i.e the lower bound. – Bunny Rabbit Dec 22 '10 at 2:19