How can i prove this asymptotic comparison?
When it comes to instances like this, I prefer to think about conceptually "promoting" and "demoting" lower order terms.

For instance, if we wanted to upper bound:

$$f(n) = 32 + 11 \log_2 n + 5 n + 2 n^2$$

We know (by intuition or limits) that $n^2$ is the fastest growing term here. Thus, we should be able to bound $f(n)$ by $n^2$. By this, I mean we can find some constant $c$ such that $f(n) \leq c n^2$ for all $n \geq n_0$ (where $n_0$ is some initial starting value for $n$ where this inequality holds true). Note, that this is preciesly the definition of Big O.


  Promoting lower order terms of $f(n)$ - Consider we can represent each lower order term by a constant $c_0$ times some function $h_0$ of $n$ (e.g. $c_0 \cdot h_0(n) = 11 \cdot \log_2 n$). For each lower order term, we will promote $h_0(n)$ to the highest order function of $n$, call it $h^*(n)$ and leave the constant the same.


Consider the lower order terms of our example: $\{32,\ 11 \log_2 n,\ 5n\}$ and the highest order term $2 n^2$. The promotions will go as follows:

$$\begin{align*}
32 &amp; \mapsto 32 \cdot n^2\\
11 \cdot \log_2 n &amp; \mapsto 11 \cdot n^2\\
5 \cdot n &amp; \mapsto 5 \cdot n^2
\end{align*}$$

Let's assume $n_0$ is the first value of $n$ for which all lower order functions of $n$ are less than or equal to the highest order function of $n$. In our example we have:

$$\begin{align*}
1 &amp; \leq n^2 &amp; \forall n \geq 1\\
\log_2 n &amp; \leq n^2 &amp; \forall n \geq 1\\
n &amp; \leq n^2 &amp; \forall n \geq 1
\end{align*}$$

Thus we have found our $n_0 = 1$. Now we can replace the lower order terms in $f(n)$ with their promoted counterparts to get:

$$\begin{align*}
f(n) &amp; = 32 + 11 \log_2 n + 5 n + 2 n^2\\
&amp; \leq 32 n^2 + 11 n^2 + 5 n^2 + 2n^2
&amp; = 50 n^2
\end{align*}$$

With this he have found our $c = 50$. Thus we can conclude:

$$f(n) \leq 50 n^2 \quad \forall n \geq 1 \quad \implies f(n) = O(n^2)$$


  For demoting lower order terms we simply remove them from the equation to get:


$$\begin{align*}
f(n) &amp; = 32 + 11 \log_2 n + 5 n + 2 n^2\\
&amp; \geq 2n^2\\
&amp;  = 2n^2
\end{align*}$$

Thus we can conclude:

$$f(n) \geq 2 n^2 \quad \forall n \geq 1 \quad \implies f(n) = \Omega(n^2)$$

Then we clearly get:

$$f(n) = O(n^2) \land f(n) = \Omega(n^2) \implies f(n) = \Theta(n^2)$$



Typically, most of these operations are simple enough to just conclude $f(n) = \Theta(n^2)$ in one fatal sweep, though it does always help to determine $c$ and $n_0$ explicitly when you need to. Clearly, the $c$ we find here is a loose upper bound and you could usually restrict it further if you needed to.

At this point I am assuming you have a summation of functions that are monotone non-decreasing. You must be a bit more careful when you have decreasing lower order terms, but these are usually not a big problem.