Prove that $\ln{n} \lt \sqrt{n}$ for $n \in \mathbb{N}^{*}$ without derivative

Question

Prove that $\ln{n} \lt \sqrt{n}$ for $n \in \mathbb{N}^{*}$.

I have tried to prove that using induction but I really don't know how to do it.

I know how to solve it by creating a function $f(x) = \sqrt{x}-\ln{x}$ and then making the derivative, and then show that is bigger than 0. But I need it solved without math analysis (without derrivative, maybe with induction).

How can I do this? Thank you very much!

Answer 1

Note that $$\ln(x) < \log_2(x).$$ now, it suffices to show that $$\log_2(x) < \sqrt{x} \Leftrightarrow x<2^{\sqrt{x}}$$ substituting $x \mapsto n^2$ we get $$n^2<2^n,$$ which is easily proven by induction for $n \ge 3$. $n = 1, 2$ are trivial.

Update:

As Del pointed out, this answer is wrong. I tried to delete it, but an accepted answer cannot be deleted. It's funny how it is accepted :).

Answer 2

Check by hand that it's true for $n\le 4$. Then show that $$\ln (n+1)-\ln(n)\le \sqrt{n+1}-\sqrt n$$ for $n\geq 5$, from which the result follows by induction. This is equivalent to $$ \ln\left(1+\frac1n\right)\le \frac{1}{\sqrt n+\sqrt{n+1}}.$$ If you know that $$ \left(1+\frac1n\right)^n\le e$$ then $$\ln\left(1+\frac1n\right)\le \frac1n \le \frac{1}{2\sqrt{n+1}}\leq \frac{1}{\sqrt n+\sqrt{n+1}}$$ because if $n\geq 5$ then $\frac1n \le \frac{1}{2\sqrt{n+1}}$.

Answer 3

While providing an answer to another question I proved this inequality $$\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1\leq \frac{n-1}{(2n-1)\sqrt{n}+1}<\frac{1}{2\sqrt{n}}$$ or $$\frac{\ln{n}}{2n}\leq \frac{1}{2\sqrt{n}}$$