Prove that $n! \geq n^{\frac{n}{2}}$

Question

I was wondering how to prove that $$n! \geq n^{\frac{n}{2}}\quad\forall n \geq 1$$ without analytic methods that relies on asympotic comparison or use of logarithms/exponentials.

Trying by induction I was stuck immediately after the induction hypothesis because I was unable to give an estimate of the following :

$$(n+1)! = n!(n+1) \geq n^{\frac{n}{2}}(n+1)$$

Is there anything I'm missing ? Any help of tip would be appreciated, as well as other methods that don't rely on seeing the inequality as an analysis task,

I'm seeking for ''arithmetic'' or ''algebraic'' demonstrations.

Answer 1

For $n \ge k\ge 1$, note that *$$k(n-k+1)-n=(n-k)(k-1) \ge 0 \implies k(n-k+1) \ge n~~~(*)$$ Next, we write $$n!=1 \cdot 2\cdot 3\cdot 4 \cdots k\cdots n ~~~(1)$$ and $$n!=n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdots (n-k+1)\cdots 1 ~~~(2).$$ Multiplying (1) and (2) pairwise as $$(n!)^2=(1\cdot n)\cdot(2\cdot (n-1))\cdot (3\cdot (n-2))\cdot (4 \cdot (n-4))\cdots (k(n-k+1))\cdots (n\cdot 1)$$ Fo from $(*)$, it follows that $$(n!)^2 \ge n^n.$$

Answer 2

I think you'll need one asymptotic comparison involving a famous base of exponentials, namely $\left(1+\frac1n\right)^n\le e$. (For what I'm about to say, you only need an upper bound of $3$, which is famously easier to prove.) The ratio$$\frac{(n+1)^{(n+1)/2}}{n^{n/2}}=\left(1+\frac1n\right)^{n/2}\sqrt{n+1}\le\sqrt{e(n+1)}$$is $\le n+1$ provided $n\ge2$ (so that $n+1>e$). So you only need check the cases $n\in\{1,\,2\}$, viz. the inequalities $1!\ge1^{1/2},\,2!\ge2^{2/2}$.

Answer 3

To show that $(n+1)!\ge(n+1)^{(n+1)/2}$, it suffices to show that \begin{align}n^{n/2}(n+1)\ge(n+1)^{(n+1)/2}&\implies n^n(n+1)^2\ge(n+1)^{n+1}\\&\implies n+1\ge\left(\frac{n+1}n\right)^n\end{align} which is true for all $n\ge2$ as the RHS is never greater than $e$. To check that case $n=1$, note that $1!\ge1^{1/2}$.

Answer 4

If so, you have to prove that $$(n+1)!\geq (n+1)^{\frac{n+1}{2}}$$ We have $$n!(n+1)\geq n^{\frac{n}{2}}(n+1)\geq (n+1)^{(n+1)/2}$$ this is equivalent $$n^{n/2}\geq (n+1)^{(n-1)/2}$$ or $$n^n\geq (n+1)^{n-1}$$ or $$n+1\geq\left(\frac{n+1}{2}\right)^n$$ which is true.