How to prove whether the statement is true or not: $$\frac{\sqrt{n}}{\log_2n} = O(n^{1/3})$$?
I know for a fact that the statement is false.
The prove doesn't have to be rigorous, I simply have to convince someone of the validity of the statement.
This is what I tried:
$$\frac{\sqrt{n}}{\log (n)}=n^{(1 / 3)}$$ $$n^{1/2}=n^{(1 / 3)} \log _{2}(n)$$
At this point, I guess it is clear to see that the left side grows faster than the right side. Thus the statement is false.
Now, could I have proved it in some other way?
Suppose that $\frac{\sqrt{n}}{\log_2(n)}=O\!\left(n^{1/3}\right)$. Then there is a $K$ so that $$ \frac{\sqrt{n}}{\log_2(n)}\le Kn^{1/3}\tag1 $$ Then, because $\log(n)\lt n$ for all $n\gt0$, $$ \begin{align} n^{1/6} &\le\frac{K}{\log(2)}\,\log\left(n\right)\\ &=\frac{12K}{\log(2)}\,\log\left(n^{1/12}\right)\\ &\le\frac{12K}{\log(2)}\,n^{1/12}\tag2 \end{align} $$ Inequality $(2)$ implies that $$ n\le\left(\frac{12K}{\log(2)}\right)^{12}\tag3 $$ which cannot be true for all $n$. Contradiction.
Therefore, $\frac{\sqrt{n}}{\log_2(n)}\not\in O\!\left(n^{1/3}\right)$.