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I am trying to show for every c, there exists $M\text{ such that }(x,y,z)\geq M$ then $S_1(x,y,z) + S_2(x,y,z) \leq ( f (x,y,z))^{-c} $ . For a particular $S_1,S_2,f$. Does it suffice to prove there exists M1, for corresponding $S_1$, with $S_1(x,y,z) \leq (f(x,y,z))^{-c} $ and $M_2$ for $S_2({x,y,z)} \leq (f(x,y,z) )^{-c} $?

This is in the same spirit as showing $O(S_1+S_2) \leq T$ can be done by showing individual upper bounds.

All functions are strictly positive increasing functions. In particular, $f$ is not a constant.

I am trying to show that there for large enough $x,y,z$, $S_1 + S_2$ is smaller than the inverse of any polynomial in $f$.

I am looking to solve it for a particular case of $f=x^2 y^2+xz\ log (xz)$ but general answers are also appreciated.

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  • $\begingroup$ If $$S_1, S_2, f^{-c} = 1$$, your weaker version holds but the proof fails. $\endgroup$ – D. Ben Knoble Sep 14 at 14:45
  • $\begingroup$ It doesn't suffice. Consider $f(x,y,z) = 1$. If you show $S_1,S_2 \leq 1$, does it follow that $S_1+S_2 \leq 1$? $\endgroup$ – Yuval Filmus Sep 14 at 14:54
  • $\begingroup$ Also, if $f(x,y,z) > 1$ and $S \leq f(x,y,z)^{-c}$ for all $c$, then $S \leq 0$. $\endgroup$ – Yuval Filmus Sep 14 at 14:55
  • $\begingroup$ @Yuval Filmus what when neither of them are constants. For a particular problem I'm looking at, I am trying to show that S1 + S2 is smaller than inverse of every polynomial in f under the proposed assumptions. $\endgroup$ – Root Sep 14 at 15:04
  • $\begingroup$ Perhaps you should tell us more about your actual problem. $\endgroup$ – Yuval Filmus Sep 14 at 16:00
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Let's consider the simpler case in which $f$ only has one input — the general case is the same, given that you use the correct definitions.

Here is what we want to prove:

Suppose that $\lim_{n\to\infty} f(n) = \infty$.

If for all $c \geq 0$ there exists $N$ such that for all $n \geq N$, $S_1(n) \leq f(n)^{-c}$ and $S_2(n) \leq f(n)^{-c}$, then for all $c \geq 0$ there exists $N$ such that for all $n \geq N$, $S_1(n) + S_2(n) \leq f(n)^{-c}$.

Let us be given $c \geq 0$. Let $N_1$ be such that $f(n) \geq 2$ for all $n \geq N_1$, and let $N_2$ be such that for all $n \geq N_2$, $S_1(n),S_2(n) \leq f(n)^{-(c+1)}$. Let $N = \max(N_1,N_2)$. Then for $n \geq N$, $$ S_1(n) + S_2(n) \leq 2f(n)^{-(c+1)} = \frac{2}{f(n)} f(n)^{-c} \leq f(n)^{-c}. $$

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