How to prove $\log n \leq \sqrt n$ over natural numbers? It seems like $$\log n \leq \sqrt n \quad \forall n \in \mathbb{N} .$$ I've tried to prove this by induction where I use $$ \log p + \log q \leq \sqrt p \sqrt q $$ when $n=pq$, but this fails for prime numbers. Does anyone know a proof?

Here is a proof of a somewhat weaker inequality that does not use calculus:
Put $m:=\lceil\sqrt{n}\>\rceil$. The set $\{2^0,2^1,\ldots,2^{m-1}\}$ is a subset of the set $\{1,2,\ldots,2^{m-1}\}$; therefore we have the inequality $m\leq 2^{m-1}$ for all $m\geq1$. It follows that
$$\log n=2\log\sqrt{n}\leq 2\log m\leq 2(m-1)\log2\leq 2\log2\>\sqrt{n}\ ,$$
where $2\log2\doteq1.386$.
