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?

That's the same as $n \le e^{\sqrt n}$ or $n^2 \le e^n$.
If we allow the power series for $e^x$, 
$e^n > n^3/6$ so $e^n > n^2$ for $n \ge 6$.
If we don't allow the power series, we can instead prove by induction that
$n^2 < 2^n$ (which, of course, is better) for $n \ge 5$:
True for $n = 5$; if true for $n \ge 5$,
$$\frac{(n+1)^2}{2^{n+1}} = \frac{n^2}{2^n}\frac{(1+1/n)^2}{2} \le (6/5)^2/2 = 36/50 < 1.$$
