Proving that $S_1+S_2 \leq f^{-\omega(1)}$ 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.
Let's consider the simpler case in which $f$ only has one input &mdash; 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 $n \geq N$,
$$
S_1(n) + S_2(n) \leq 2f(n)^{-(c+1)} = \frac{2}{f(n)} f(n)^{-c} \leq f(n)^{-c}.
$$