Show that $O(\text{max}\{f(n),g(n)\})=O(f(n)+g(n))$

Show that $ O(\text{max}\{f(n),g(n)\})=O(f(n)+g(n))$

Can I keep the same constant $ c$ in each of the cases?

Consider two cases: $ $ 1)f(n)>g(n);O(\text{max}\{f(n),g(n)\})⇒O(f(n))\Rightarrow d(n) ≤c⋅f(n);c>0$ $ $ $ 2)f(n)≤g(n);O(\text{max}\{f(n),g(n)\})⇒O(g(n))\Rightarrow e(n)≤c⋅g(n);c>0$ $ Combining the 2 cases yields: $ $ d(n)+e(n)≤c⋅f(n)+c⋅g(n)$ $ $ $ d(n)+e(n)≤c⋅(f(n)+g(n))$ $ Which is definition of $ O(f(n)+g(n))$