I'm having trouble forming a mathematical proof for a question. I can write down thousands of examples with various values of n that shows it's correct, but I'm not sure how to turn that into a mathematical proof it when multiple functions are involved.
$f_{1}, f_{2}$ and $f_{3}$ are functions on positive real numbers. Prove that if $f_{1}(n) \in O(f_{2}(n))$ and $f_{2}(n) \in O(f_{3}(n))$ then $f_{1}(n) \times f_{2}(n) \in O(f_{3}(n)^{2})$
On
So we have three functions, all of them defined (and positive, for simplicity) for sufficiently large $x\in{\mathbb R}$.
The statement $f_1(x)=O\bigl(f_2(x)\bigr)$ means the follwing: There is a $C_1$ and an $x_1$ such that
$$f_1(x)\leq C_1 f_2(x)\qquad(x>x_1)\ .$$ Similarly, $f_2(x)=O\bigl(f_3(x)\bigr)$ means that there is a $C_2$ and an $x_2$ such that $$f_2(x)\leq C_2 f_3(x)\qquad(x>x_2)\ .$$ Then for $x>\max\{x_1,x_2\}$ we have $$f_1(x)\>f_2(x)\leq C_1 f_2(x)\cdot C_2f_3(x)\leq C_1C_2 f_3(x)\cdot C_2 f_3(x)=C\bigr(f_3(x)\bigr)^2$$ with $C:=C_1C_2^2$. This proves the claim.
Hint:
$$\begin{cases}\left|\frac{f_1(n)}{f_2(n)}\right|\le M_1\\{}\\\left|\frac{f_2(n)}{f_3(n)}\right|\le M_2\end{cases}\;\;\implies \left|\frac{f_1(n)f_2(n)}{f_3(n)^2}\right|=\left|\frac{f_1(n)}{f_3(n)}\right|\left|\frac{f_2(n)}{f_3(n)}\right|\le \ldots\ldots?$$