3

1°

$n_0=1$

$(1!)^2 \ge 1^1$

$1\ge1$

2°

$k \ge n_0$

assumption: $$(k!)^2 \ge k^k$$

and for k+1: $$((k+1)!)^2 \ge (k+1)^{k+1}$$

I also noticed that: $$((k+1)!)^2 = (k!)^2 * (k+1)^2$$

Overkillus
  • 104

2 Answers2

1

A direct approach may be simpler: We have $$(n!)^2=\prod_{k=1}^nk\cdot\prod_{k=1}^{n}(n+1-k)=\prod_{k=1}^nk(n+1-k)\ge\prod_{k=1}^nn $$ because $k(n+1-k)=-k^2+(n+1)k$ is a "downward" quadratic, hence is minimal at the boundary places $k=1$ and $k=n$.

Hagen von Eitzen
  • 374,180
1

Continuing from your steps, we are just required to prove

$$k^k \geq (k+1)^{k-1} \Leftrightarrow k^{k-1} \cdot k \geq (k+1)^{k-1}\Leftrightarrow (\frac{k}{k+1})^{k-1} \geq \frac{1}{k}$$

Using Bernouli's inequality, we have

$$(\frac{k}{k+1})^{k-1}=(1-\frac{1}{k+1})^{k-1} \geq 1-\frac{k-1}{k+1}=\frac{2}{k+1}$$

So, we are just left to prove $$\frac{2}{k+1} \geq \frac{1}{k}\Rightarrow 2k \geq (k+1) \Rightarrow k \geq 1$$ which is obvious.

Roby5
  • 4,287
  • Sorry, but I dont really understand this way. From where did you get: $(\frac{k}{k+1})^{k-1} \geq \frac{1}{k}$ – Overkillus Jun 12 '16 at 22:22
  • I had a day off and just went back to it. Still i try to come up with an idea how did you get this part. On my own I came to: $n >= ((n+1)/n)^(n-1) – Overkillus Jun 14 '16 at 08:20
  • 1
    @Overkillus What's the confusion? $n \geq (\frac{n+1}{n})^{n-1} \Rightarrow (\frac{n}{n+1})^{n-1} \geq \frac{1}{n}$ – Roby5 Jun 14 '16 at 09:09