1

First I've checked it by putting $n=3$.

Then, I've assumed it to be true for $n=k$ to get $k!>k^{\frac k2}$.

Then, putting $n=k+1$, I get $(k+1)!=k!(k+1)>k^{\frac k2}(k+1)$. Now, I'm stuck and couldn't proceed further.

Parcly Taxel
  • 103,344
ankit
  • 2,381

3 Answers3

4

It remains to show that $(k^{k/2})(k+1)\geq (k+1)^{(k+1)/2}$, or $k^k\geq (k+1)^{k-1}$. This is equivalent to $$k\geq\left(1+\frac1k\right)^{k-1}.$$ Does the term on the right look familiar?

pi66
  • 7,164
1

Consider an integer $n\gt2.$ Assuming that $m!\gt m^{m/2}$ holds whenever $2\lt m\lt n,$ I have to show that $$n!\gt n^{n/2}$$ or equivalently that $$(n!)^2\gt n^n$$ or equivalently that $$\prod_{x=1}^n x(n+1-x)\gt\prod_{x=1}^n n.$$ Note that $x(n+1-x)\ge n;$ the absolute minimum value of the function $f(x)=x(n+1-x)$ on $[1,n]$ is $f(1)=f(n)=n.$ This shows that $$\prod_{x=1}^n x(n+1-x)\ge\prod_{x=1}^n n.$$ To see that the inequality is strict, note that, since $n\gt2,$ for $x=2$ we have $x(n+1-x)=2n-2\gt n.$ Q.E.D.

(There's no law that a proof by induction has to use the inductive hypothesis, is there?)

bof
  • 78,265
-1

It's an easy fact that $${k \choose m} \leq k^m.$$ Then $$(k+1)^{k+1} = k^{k+1} + {k \choose 1}k^k + {k \choose 2}k^{k-1}+ \cdots +1 \leq k^{k+1}+\cdots + k^{k+1} = k^{k+1}(k+1)$$ since there are $k+1$ terms in the sum. If we take the square root of both sides, this gives your last inequality, so apply the induction hypothesis and you're done.

For the "easy fact", note that $${k \choose m} = \frac{k(k-1)\cdots(k-m+1)}{(k-m)!} < \frac{k^m}{(k-m)!} \leq k^m.$$

B. Goddard
  • 1