Proof for $(n!)^2 \geq n^n$ without using induction [duplicate]

Question

The question is to prove that $$ (n!)^2 \geq n^n \text{ | } n\in\mathbb{N} $$

My attempt:

From the general AM-GM inequality, we have $$ \frac{a_1+a_2+a_3+\cdots+a_n}{n} \geq \sqrt[n]{a_1a_2a_3\cdots a_n} $$

Taking $$ a_r=\frac{1}{r^2} $$ we get

$$ \begin{aligned} &\left(\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}{n}\right)^n > \left(\frac{1}{1.2^2.3^2\cdots n^2}\right) \\ \implies & \left(\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}{n}\right)^n > \frac{1}{(n!)^2} \\ \implies & (n!)^2 > \left(\frac{n}{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}\right)^n \end{aligned} $$ If $$ \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)^n=k $$ Then, we have $$ (n!)^2 > \frac{n^n}{k} $$ and I'm stuck. Help please!

Answer 1

Hint:

$$(n!)^2=(1\cdot n)\cdot(2\cdot (n-1))\cdots((n-1)\cdot 2)\cdot(n\cdot 1)$$

Then show:

$$k\cdot (n+1-k)\geq n$$

for $k=1,\dots,n,$ with equality only when $k=1$ or $k=n.$

Answer 2

I assume you are talking about $n\geq3$, otherwise your proposition is false. Your proof could be done by induction. The base case, $n=3$, is trivially shown. For the inductive step, let us consider the functions: $$f(k)=\left(1+\frac 1k\right)^k$$ and $$g(k)=k+1$$ Try to show that for all $k\geq 3$, $g(k)>f(k)$. For this, you can use calculus. Notice that both $f$ and $g$ are monotonically increasing and both $g(3)>f(3)$ and $\lim_{k\to \infty}g(k)>\lim_{k\to \infty}f(k)$. Once you've proven this, if we take a statement $S_k$ as: "$(k!)^2>k^k$". Then, $$k+1>\left(1+\frac 1k\right)^k$$ So, $$(k+1)k^k>(k+1)^k$$ So, $$k^k>(k+1)^{k-1}...(1)$$ Which means, $$(k+1)^2 k^k>(k+1)^{k+1}$$ Now, multiplying $(k+1)^2$ on both sides of $S_k$, we get: $$((k+1)!)^2>(k+1)^2 k^k...(2)$$ Combining $(1)$ and $(2)$, you get that $S_{k+1}$ is also true. QED.