Finding $c$ such that $c\sqrt{n} \geq \log_2 n$ for all $n\geq 2$.

Question

I am struggling with the follwing problem:

It is clear to me, that $\sqrt{n}$ is larger than $\log_2 n$ for all $n\geq17$. This is not the case for all $n\leq16$ though, since $\sqrt{9} = 3 < \log_2 9$ for example.

How can I find a minimal constant $c\in \mathbb{R}_{>0}$, such that $c\sqrt{n} \geq \log_2 n$ for all $n\geq 2$?

I would appreciate, if one of you has a good idea, how to solve this, thank you.

Answer 1

Define $$f(n)\equiv\frac{\log_2 n}{\sqrt{n}}\quad\text{for all $n\geq 2$.}$$ Its derivative is $$f'(n)=\frac{2-\ln n}{\sqrt{n^3}\ln 4},$$ where $\ln$ denotes the natural logarithm. Therefore, $f'(n)\gtrless0$ according as $n\lessgtr e^2$, and $f$ reaches its unique maximum at $n=e^2$ on the interval $[2,\infty)$. Consequently, $$f(e^2)=\frac{2}{e\ln 2}$$ is the least $c$ satisfying the desired inequality.