If we know: $$f_1, f_2, g_1, g_2, c_1, c_2 > 0$$ $$f_1( n ) \leq c_1 g_1( n ) ~\forall n \geq n_1$$ $$f_2(n) \leq c_2 g_2( n ) ~\forall n \geq n_2$$ Then: $$f_1(n)f_2(n) \leq c_1c_2g_1(n)g_2(n)$$ but for what n? An example that I did yielded $n = \sqrt{n_1n_2}$ but I cannot prove why this is or if it is general.

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    $\begingroup$ To make your final conclusion about the product of functions, you need to use the previous two inequalities in a regime where they are both valid. So you just need $n \geq \max\{n_1, n_2\}$. If $n<\max\{n_1,n_2\}$, at least one of the required prior inequalities is no longer guaranteed, and no conclusions whatsoever can be made for such cases. $\endgroup$ – Michael Mar 15 at 22:11
  • $\begingroup$ Since we want to use both of the original bounds, we need $n$ to lie both in the set $\{n_{1}, n_{1} + 1, \ldots \}$ and in the set $\{n_{2}, n_{2} + 1, \ldots \}$. Therefore, for all $n \geq $ (the largest of $n_{1}, n_{2}$). $\endgroup$ – avs Mar 15 at 22:15

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