Qusetion about Gamma function ※I'm not good at English, so please edit my question※
I read a question ,
and tried to prove $\Gamma(x) \geq x^{3}$ For all real number $x \geq 6$ using gamma function's definition and its derivative , but I couldn't prove inequality $\Gamma(x) \geq x^{3}$
This is my attempt.

For $ n \in  \mathbb{N} $, $ n! =  \Gamma \left ( n + 1 \right )$ 
Gamma Function's definiton is $\Gamma \left ( x \right ) = \int_{0}^{\infty }t^{x-1}e^{-t} \mathrm{d}t$
Let $f\left ( x , a \right ) = \int_{0}^{a}t^{x-1}e^{-t} \mathrm{d}t$
(Then $\lim_{a \to \infty} f\left ( x,a \right ) = \Gamma (x)$)
$\Rightarrow f\left ( x,a \right ) = \lim_{n \to \infty}\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}$
For partial sum$\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}$ , $ 0 \leq  \frac{k}{n} \leq 1$ Because  $ n \geq k \geq 1$ 
Let $ K = \frac{ka}{n}$ ( $ a \geq K \geq 0 $)
Then $\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}= \frac{a}{n}\sum_{k=1}^{n} K^{x-1}e^{-K}$
Let $ 0 < a < n$ ,  $ 1 > C = \frac{a}{n}$ ($K=kC$)
We get$\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}= C^{x} \sum_{k=1}^{n}k^{x-1}e^{-Ck}$

But, I think that isn't proper approaching
How proof inequality $\Gamma(x) \geq x^{3}$  using gamma function's definition?
I don't have any idea for proving this

1) Statement is false as stated:
$$\Gamma\left(6\right) = 5!=120$$
$$6^{3} = 216$$
2) Proposition: $\Gamma\left(x\right)\geq x^{3}$
  for all real $x\geq7$. 
Proof:
$$\frac{\Gamma\left(x\right)}{x^{3}}=\frac{\left(x-1\right)\Gamma\left(x-1\right)}{x^{3}}=...=\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}{x^{3}}\Gamma\left(x-4\right)$$
 Then, since $\Gamma\left(x-4\right)\geq1$
  for $x\geq5$: $$\frac{\Gamma\left(x\right)}{x^{3}}\geq\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}{x^{3}},\textrm{ }\forall x\geq5$$
Next: $$\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}{x^{3}}=\left(1-\frac{1}{x}\right)\left(1-\frac{2}{x}\right)\left(1-\frac{3}{x}\right)\left(x-4\right)$$
 Then, $x\geq7$ implies $-\frac{1}{x}\geq-\frac{1}{7}$.
Thus:
$$\left(1-\frac{1}{x}\right)\left(1-\frac{2}{x}\right)\left(1-\frac{3}{x}\right)\left(x-4\right)\geq\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{3}{7}\right)\left(7-4\right)=\frac{360}{343}\geq1$$
Thus, $x\geq7$ implies $$\frac{\Gamma\left(x\right)}{x^{3}}\geq1$$.
Q.E.D.
