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?

Bigger by a Factor
Using $e^x\ge1+x$, and substituting $e^x=\sqrt{u}$ and $e^2u=n$, we get
$$
\begin{align}
e^x\ge1+x
&\implies\sqrt{u}\ge1+\frac12\log(u)=\frac12\log(e^2u)\\
&\implies\frac2e\sqrt{n}\ge\log(n)
\end{align}
$$
which is stronger than $\sqrt{n}\ge\log(n)$.

Bigger by an Offset
we could also set $4u=n$, which gives
$$
\sqrt{u}\ge1+\frac12\log(u)
\implies\sqrt{n}\ge2(1-\log(2))+\log(n)
$$
which is also stronger than $\sqrt{n}\ge\log(n)$.
