TITLE: Prove that $\sqrt{n} > \ln n$
QUESTION [7 upvotes]: Prove that $\sqrt{n} > \ln n$ for all $n \in \mathbb{N}$.
I need to use this fact for one of the proofs that I am working on. However, I am having trouble proving this. I tried induction but don't really know how to do it. Can someone help me with this?

REPLY [4 votes]: For any $n\geq 1$, by the Cauchy-Schwarz inequality ($\text{CS}$),
$$ \log n=\int_{1}^{n}\frac{dx}{x}\color{red}{\stackrel{\text{CS}}{\leq}}\sqrt{\int_{1}^{n}dx\int_{1}^{n}\frac{dx}{x^2}}=\sqrt{(n-1)\left(1-\tfrac{1}{n}\right)}=\sqrt{n}-\frac{1}{\sqrt{n}}.$$