Prove that $\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $ 
$$\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $$

I tried it prove it by mathematical induction but failed  .
For $n=6$
$$(3)^6 \gt 6!$$
Now for $n=k$
$$\left(\frac{k}{2}\right)^k \gt k!$$
Now for $n=k+1$
$$\left(\frac{(k+1)}{2}\right)^{k+1} \gt (k+1)!$$
$$\left(\frac{(k+1)}{2}\right)^{k} \gt 2(k!)$$
 A: METHODOLOGY $1$:  Proof by Induction 

PRIMER:
In THIS ANSWER, I showed using Bernoulli's Inequality that $\left(1+\frac1n\right)^n$ is a monotonically increasing sequence.  Therefore, its minimum is $2$ when $n=1$.  It is clearly bounded by $3$ as can be shown applying the binomial theorem and applying simple estimates.
$$\begin{align}\left(1+\frac1n\right)^n&=1+1\\\\&+\frac1{2!}\left(1-\frac1n\right)+\frac{1}{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots +\frac{1}{n!}\left(1-\frac1n\right)\cdots\left(1-\frac{n-1}n\right)\\\\&\le 1+1+\frac12+\frac1{2^2}+\cdots +\frac{1}{2^{n-1}}\\\\&\le 3\end{align}$$

Now, assume for some $n>6$ that $\left(\frac{n}{2}\right)^n\ge n!\ge \left(\frac{n}{3}\right)^n$.  Then, for $n+1$ we have
$$\begin{align}
\left(\frac{n+1}{2}\right)^{n+1}&=\left(\frac{n+1}{2}\right)\,\left(\frac{n}{2}\right)^n\,\left(1+\frac1n\right)^n\\\\
&\ge \left(\frac{n+1}{2}\right)\,n!\,\left(1+\frac1n\right)^n\\\\
&=(n+1)!\,\frac12 \,\left(1+\frac1n\right)^n\\\\
&\ge (n+1)!
\end{align}$$
as was to be shown, and 
$$\begin{align}
\left(\frac{n+1}{3}\right)^{n+1}&=\left(\frac{n+1}{3}\right)\,\left(\frac{n}{3}\right)^n\,\left(1+\frac1n\right)^n\\\\
&\le \left(\frac{n+1}{3}\right)\,n!\,\left(1+\frac1n\right)^n\\\\
&=(n+1)!\,\frac13 \,\left(1+\frac1n\right)^n\\\\
&\le (n+1)!
\end{align}$$
as was to be shown!

METHODOLOGY $2$:  Proof Invoking Bounds of Stirling Formula
We can use the bounds found from development of Stirling's Formula (See "Advanced Calculus, John Olmsted, 1961, pp. 490-491.)
$$\left(1+\frac{1}{12(n+1)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\le n!\le \left(1+\frac{1}{12(n-2)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \tag 1$$
for $n\ge 3$.
For the left-hand side of $(1)$, we write 
$$\begin{align}
n!&\ge \left(1+\frac{1}{12(n+1)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\\\
&=\color{blue}{\left(1+\frac{1}{12(n+1)}\right)}\color{green}{\sqrt{2\pi n}\left(\frac{3}{e}\right)^n}\left(\frac{n}{3}\right)^n\\\\
&\ge \color{blue}{(1)}\,\color{green}{(1)}\,\left(\frac{n}{3}\right)^n\\\\
&=\left(\frac{n}{3}\right)^n
\end{align}$$
For the right-hand side of $(1)$, we write 
$$\begin{align}
n!&\le \left(1+\frac{1}{12(n-2)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\\\
 &=\color{blue}{\left(1+\frac{1}{12(n-2)}\right)}\color{green}{\sqrt{2\pi n}\,\left(\frac{2}{n}\right)^n}
\,\left(\frac{n}{2}\right)^n
\end{align}$$
The function $f(x)=\sqrt{2\pi x}\left(\frac{3}{e}\right)^x$ can easily be shown to monotonically decrease for $x>\frac{1}{2\log(e/2)}$.  Therefore, for $x=n\ge 6$, the function is less than $f(6)<\frac{48}{49}$.  And inasmuch as $ 1+\frac{1}{12(n-2)}\le \frac{49}{48}$ for $n\ge 6$, we have
$$\begin{align}
n!&\le \left(1+\frac{1}{12(n-2)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\\\
 &=\color{blue}{\left(1+\frac{1}{12(n-2)}\right)}\color{green}{\sqrt{2\pi n}\,\left(\frac{2}{n}\right)^n}
\,\left(\frac{n}{2}\right)^n\\\\
&\le \color{blue}{\left(\frac{48}{49}\right)}\,\color{green}{\left(\frac{49}{48}\right)}\,\left(\frac{n}{2}\right)^n
\end{align}$$
And we are done!
A: $$\begin{align}
\left(\frac{k+1}2\right)^{k+1}&=\frac {k+1}2\left(\frac {k+1}2\right)^k\\
&>\frac{k+1}2\left[\left(\frac k2\right)^k+\binom k1\left(\frac k2\right)^{k-1}\frac12\right]\\
&>\frac{k+1}2[k!+k!]\\
&=(k+1)!
\end{align}$$
A: Still another approach is to exploit creative telescoping. We have
$$ n = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) \tag{1}$$
hence it follows that
$$ n! = \frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}\in\left(\frac{n^n}{e^{n-1}},\frac{n^n}{2(9/4)^{n-2}}\right)\tag{2} $$
for any $n\geq 2$.
