Some simple inequalities on factorials How can I prove the following inequalities for factorials in an elemantary way:
$$ n^{\frac{n}{2}} \leq n! $$
$$ n^{\frac{2n}{3}} \leq n! $$
$$ \left(\frac{n}{3}\right)^{n} \leq n! $$
Usage of stirling approximation is not allowed. Thank you.
Edit: You can show the truth of the second inequality for $n>k$ with a information that contains what is $k$.
 A: Point 1. Fix $n$, then
$$
(n!)^2=(1\cdot n) \cdot (2\cdot (n-1)) \cdot (3 \cdot (n-3)) \cdots (n \cdot 1)
$$
and each factor in bracket is at least $n$.
Point 3. Fix $n=3k$ (the other cases can be done similarly), then
\begin{align}
(3k)!&\ge (3k)(3k-1) \cdots (k) \\
&= (2k)\prod_{i=1}^k (2k+i)(2k-i)\\
&\ge (2k)\prod_{i=1}^k (2k+k)(2k-k)\\
&=(2k)(3k^2)^k= (2\cdot 3^k) k^{3k}.
\end{align}
A: Inequality 2 holds for $n \ge 12$ (but not before that except $n=1.$)
First let $a_n=n^{2n},\ b_n=(n!)^3,$ so you want $a_n \le b_n.$
For $n=12:$ $a_{12}=0794....,\ b_{12}=1099...,$ where the sections after the dots have the same number of digits.
Now $a_{n+1}/a_n=(1+1/n)^{2n} \cdot (n+1)^2,$ while $b_{n+1}/b_n=(n+1)^3.$ Here $(1+1/n)^{2n} \to e^2$ from below, so at any rate is less than $8.$ Since we're working with $n \ge 12$ the inductive step wanted is now $8(n+1)^2 \le (n+1)^3$ and holds.
A: As an alternative we can prove by induction, that is for $n!\ge n^{\frac{n}{2}}$
base case


*

*$n=1 \implies 1\le 1$


induction step
$(n+1)!=(n+1)n!=(n+1)n^{\frac{n}{2}}\stackrel{?}\ge (n+1)^{\frac{n+1}{2}}$
$(n+1)n^{\frac{n}{2}}\ge (n+1)^{\frac{n+1}{2}} \iff n^{\frac{n}{2}}\ge (n+1)^{\frac{n-1}{2}} \iff n^n\ge (n+1)^{n-1}\\\iff n\,n^{n-1}\ge (n+1)^{n-1}\iff n\ge \left(1+\frac1n\right)^{n-1}$
which is true for $n=1$, $n=2$ by inspection and $n\ge 3$ since $RHS <3$.
