# Which side do I apply AM-GM on?

$\sqrt[n]{n!} \geq \sqrt{n}$

Prove the above for all positive integers $$n$$.

• Try to give a non-inductive, non-calculus proof

Note by Sharky Kesa
2 years, 3 months ago

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## Comments

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Try to give a non-inductive, non-calculus proof

So I will give an inductive, calculus-requiring proof.

Lemma 1: $$\left( 1 + \frac{1}{n} \right)^n \le e$$ for all positive integer $$n$$.

One way to prove it is to observe that $$1 + \frac{1}{n} \le 1 + \frac{2}{2n} + \frac{1}{4n^2} = \left( 1 + \frac{1}{2n} \right)^2$$, thus by induction,

\displaystyle\begin{align*} \left(1 + \frac{1}{n} \right)^n &\le \left(1 + \frac{1}{2n} \right)^{2n} \\ &\le \left(1 + \frac{1}{4n} \right)^{4n} \\ &\le \ldots \\ &\le \lim_{k \to \infty} \left(1 + \frac{1}{2^k n} \right)^{2^k n} \\ &= \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n \\ &= e \end{align*}

Lemma 2: $$e < 3$$

\displaystyle\begin{align*} e &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \ldots \\ &< 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots \\ &= 1 + 2 \\ &= 3 \end{align*}

Now, the claim $$\sqrt[n]{n!} \ge \sqrt{n}$$ is equivalent to $$(n!)^2 \ge n^n$$. We will prove this by induction. The base cases are $$n = 1, 2, 3$$ in which it is true. Now suppose that it's true for some $$n \ge 3$$, and we'll prove the claim for $$n \gets n+1$$.

\displaystyle\begin{align*} ((n+1)!)^2 &= (n!)^2 \cdot (n+1)^2 \\ &\ge n^n \cdot (n+1)^2 \\ &\ge n^n \cdot (n+1) \cdot e & \text{by Lemma 2, because n \ge 3 > e} \\ &\ge n^n \cdot (n+1) \cdot \left( 1 + \frac{1}{n} \right)^n & \text{by Lemma 1} \\ &= \left(n + \frac{n}{n} \right)^n \cdot (n+1) \\ &= (n+1)^n \cdot (n+1) \\ &= (n+1)^{n+1} \end{align*}

- 2 years, 3 months ago

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We have to prove $$(n!)^2 \ge n^n$$. Instead of proving it using Sterling's approximation or induction here is a solution on AM-GM. I found a nice form of $$(n!)^2$$ recently. $$(n!)^2=[1 \cdot n][2 \cdot (n-1)]\ldots[n \cdot1]$$(there are $$n$$ terms). Now, for some variable $$m$$, such that $$1 \le m \le n$$ we have $$m(n-m+1) \ge n$$. The result follows.

- 2 years, 3 months ago

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Wise solution!

- 2 years, 3 months ago

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Why is it that for a variable $m$
$1\leq m\leq n$ is $m(n-m+1)\geq n$ ?

- 2 years, 3 months ago

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Take $$n=m+k$$ where $$k \ge 0$$. The inequality simplifies to $$m(k+1) \ge m+k$$ or $$mk+m \ge m+k$$ or $$m \ge 1$$ which is true.

- 2 years, 3 months ago

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OK.. Thanks!

- 2 years, 3 months ago

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Nice solution

- 2 years, 3 months ago

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$$\sqrt[n]{n!} \geq \sqrt{n}$$

$${n!} \geq {\sqrt{n}}^n$$

$$(n) \cdot (n-1) \cdot (n-2) \cdot \dots \cdot (1) \geq {\sqrt{n}}^n$$

$$(n)(1) \cdot (n-1)(1+1) \cdot (n-2)(1+2) \cdot \dots \cdot (\frac{n+1}{2})(\frac{n-1}{2}) \geq {\sqrt{n}}^n$$ or $$(n)(1) \cdot (n-1)(1+1) \cdot (n-2)(1+2) \cdot \dots \cdot (\frac{n}{2}) \geq {\sqrt{n}}^n$$

We note that (n-k)(1+k) > n for all -1<k<n+1. So the result follows. By the way, i was sniped by Svatejas Shivakumar.

- 2 years, 3 months ago

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Ninja'ed haha tfs

- 2 years, 3 months ago

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Well what about n-1?

- 2 years, 3 months ago

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