※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
