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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?

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4 Answers 4

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 $$ \sqrt n>\ln n\qquad\text{for all }n>0.$$

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how to u get to 3? why e^2x becomes e^x? –  user2675516 Apr 20 '14 at 12:44
Because since $e^{2x} > 4x^2 = 2x \cdot 2x = (2x)^2 $, then we can put $y = 2x$ and we obtain $e^y > y^2 $ –  E 110th St and 5th Ave Apr 24 '14 at 4:37
I don't understand this post, in that sense that how do you prove the "important" inequality? I am presuming you just do what Dutta's post does, and differentiate... –  user1729 Apr 28 '14 at 11:59

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.

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Do you happen to know any non-calculus way of doing this? thanks –  user2675516 Apr 20 '14 at 8:40
I do not know. Hope some other people will come with answer. –  Dutta Apr 20 '14 at 8:45
thanks anyway :) –  user2675516 Apr 20 '14 at 8:46
Actually $f$ is only increasing on $(4,+\infty)$, not on $(1,+\infty)$. Thus the argument can be saved, but it needs to be modified. // @upvoters Why the upvote? –  Did Apr 20 '14 at 9:21
@derivative Thank you for identifying it. My mistake and same mistake has done by many people. –  Dutta Apr 20 '14 at 9:36

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.

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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*}

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