How do i show that $\Gamma (x) > 1/e \int_{0}^{1} t^{x-1} dt$ when $x>0$?

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Prove that $\Gamma (x) > \frac{1}{e} \int_{0}^{1} t^{x-1} dt$ when $x>0$? I can understand $e^{-t}$ will appear in the gamma function but how do i prove this? I cannot do anything more help please!

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$\Gamma(x)=\int_{0}^{\infty} e^{-t}t^{x-1} dt=\int_{0}^{1}e^{-t}t^{x-1} dt+\int_{1}^{\infty}e^{-t}t^{x-1} > \int_{0}^{1} e^{-t}t^{x-1} dt=e^{-k}\int_{0}^{1}t^{x-1}dt$ for some $k\in (0,1)$ by the Mean value theorem, $> e^{-1}\int_{0}^{1}t^{x-1} dt$