How to show that the inequality with exponents is true

Question

In part of a question in my homework I'm doing, I need to show that:

$$ \log(n) > \sqrt [n]{n} $$

Is this a correct a correct manipulation: $$n > e^\sqrt [n]{n}$$

Thank you in advance!

edit: that's a good point that it works only for n>4, and I forgot to mention that n is a natural number ({1,2,...})

Answer 1

Well, $\log(1) > \sqrt[1]1$ is wrong, and as well n=2,3,4. So let's suppose $n \geq 5$ (it holds for $n=5$).

Put it as $$ (\log (n))^n > n $$

Then use induction, $$ %(\log (n+1))^{n+1} >=(\log (n) + \log(1 + 1/n))^{n+1} (\log (n+1))^{n+1} >(\log (n))^{n+1} = \log(n) (\log (n))^{n} \geq n \log(n) $$

Where the last one holds by induction.

Now show

$$\log(n) > 1 + 1/n$$ which holds for $n=5$, and then also for larger n since the log is increasing with n and $1/n$ is decreasing with n. So one can continue $$ %(\log (n+1))^{n+1} >=(\log (n) + \log(1 + 1/n))^{n+1} (\log (n+1))^{n+1} \geq n \log(n) > n (1 + 1/n) = n+1 $$ Done. $\quad \quad \Box$

Answer 2

(In the true spirit of stackexchange:)

Yes.

Answer 3

Here's a proof by induction since $n \in \mathbb{Z^{+}}$

$n=1$, we have:

LHS = $\log 1$ = 0

RHS = $1 > $LHS hence the statement is incorrect.

Answer 4

True, lets manipulate the inequality $ \ln(n) > n^{\frac{1}{n}}$ which can be turn into $\ln(\ln(n)) > \frac{1}{n} \ln(n)$ (because $\ln(a^b) = b \ln(a)$)

And then multiply by $n$ we get that $n \ln(\ln(n)) > \ln(n)$ when $\ln(\ln(n)) > 1$ we can reduce it to $n > \ln(n)$ which can be turn into $e^n > n$ which is true and can be proven by induction $n=1$ => $e^1 \approx 2.718 >1 $ now assume its true for $e^{k} > k$ prove that $e^{k+1} > k+1$ which is $e^{k}*e >k+1$ which is $k*e > k+1$ using the assumption,which can be reduces to $2k>k+1$ and its true when $k>1$, last step is to see when $\ln(\ln(n)) >1$ which is just when $n > e^{e^{1}} \approx 15.15$ ,so the inequality is true when $n >15$ and when $n \leq 15$ a check is enough to show that its true when $n >4$

Answer 5

Andreas said to holds for $5$ , which is not true ; Ahmad has the correct concepts but uses the $\log_e$ while the question uses base $10....n ≥16$ for the inequality to hold , then one can take the derivative of both sides and see the left increases and right decreases for $n ≥ 16 \cdots$ thus it holds for $n ≥ 16$