Prove that $\sqrt{n} > \ln n$ 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?
 A: Notice that $$\frac{\ln(n)}{\sqrt{n}} = 2 \frac{\ln(\sqrt{n})}{\sqrt{n}},$$ so it is sufficient to show that $$\frac{\ln(x)}{x}< \frac{1}{2} \ \text{for all} \ x >0 \hspace{1cm} (1)$$ 
If $f(x)= \frac{\ln(x)}{x}$ then $f'(x)= \frac{1-\ln(x)}{x^2}$, so $f$ is nondecreasing on $(0,e]$ and nonincreasing on $[e,+ \infty)$. Therefore, $f$ has a global maximum at $x=e$. Because $f(e)= \frac{1}{e} < \frac{1}{2}$, $(1)$ may be deduced.
A: let $\sqrt n > \ln n$, then 
$$\sqrt{n+1} > \sqrt{(\log n)^2-( \log(n+1))^2+( \log(n+1))^2+1}> \log (n+1) $$
hence proved by induction, that is if we show that
$$\log(n)^2 - \log(n+1)^2 + 1 > 0$$
which you can find the maximum using calculus which is less than $1$. Also you can try this,
\begin{align*}
\log(n+1)^2 - \log(n)^2 &= (\log n + \log(1 +1/n))^2 - \log(n)^2\\ 
 &= 2 \log(1 + 1/n) \log n+ \log(1+1/n)^2\\ 
 &\le 2 \left( \frac 1 n  \right ) \log n +\left( \frac 1 n \right )^2\\ 
 & < 2 \left( \frac 1 n  \right ) \sqrt n +\left( \frac 1 n \right )^2 \\
 & <1 \;  \forall n>5
\end{align*}
A: For any $n\geq 1$, by the Cauchy-Schwarz inequality ($\text{CS}$),
$$ \log n=\int_{1}^{n}\frac{dx}{x}\color{red}{\stackrel{\text{CS}}{\leq}}\sqrt{\int_{1}^{n}dx\int_{1}^{n}\frac{dx}{x^2}}=\sqrt{(n-1)\left(1-\tfrac{1}{n}\right)}=\sqrt{n}-\frac{1}{\sqrt{n}}.$$
A: The most important inequality about the exponential is
$$\tag1e^x\ge 1+x\qquad\text{for all }x\in\mathbb R\text{ with equality iff }x=0.$$
From this we find 
$$\tag2 e^{x}=(e^{x/2})^2\stackrel{(a)}\ge (1+x/2)^2=(1-x/2)^2+2x\stackrel{(b)}\ge 2x$$
where $(a)$ holds for $x\ge-2$ and is strict for $x\ne0$, and $(b)$ is strict for $x\ne 2$. Hence $e^{x}>2x$ at least for $x\ge-2$ and in fact trivially from $2x<0<e^x$ also for $x<-2$.
By squaring $e^{2x}>4x^2$ for all $x\ge 0$, i.e. 
$$\tag3 e^x>x^2\qquad\text{for all }x\ge0.$$
With $x=\sqrt n>0$ and taking logarithms on both sides we get
$$ \sqrt n>\ln n\qquad\text{for all }n>0.$$
A: Consider the function $f(x) = \sqrt x - \log x$ and differentiate:
$$f'(x) = \frac{1}{2 \sqrt x} - \frac{1}{x} = \frac{\sqrt x - 2}{2x}.
$$
Hence $f'(x) \le0$ for all $0\lt x\le4$ and $f'(x)\ge 0$ for all $x \ge 4$. So the function is monotone decreasing for all $0\lt x \le 4$ and monotone increasing for all $x \ge 4$.
Also $f(4) = \sqrt 4 - \log4=2-2\log2=2\log(\mathrm e/2) > 0$.
Thus $f(x) > 0$ for all $x \gt0$, that is, $\sqrt x - \log x > 0$.
Now put $x = n $ and get your result.
