Prove that $\Gamma(x)\geq x-1$ for $x \geq 3$.
We have defined $\Gamma(x)=\int _{0}^{\infty}t^{x-1}e^{-t}dt$, where $x>0$.
This is a review problem for an exam please give a full complete solution.
Thanks.
Update: I have completed a proof by induction and will post it later.
On
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Prove that $\ds{\Gamma\pars{x} \geq x - 1}$ for $\ds{x \geq 3}$. $\bbox[15px,border:1px dotted navy]{\ds{\mbox{It's}\ sufficient\ \mbox{to prove that}\ \Gamma\pars{x} \geq 1\ \mbox{for}\ x \geq 2}}$.
\begin{align} \Gamma\pars{x} & = \int_{0}^{\infty}t^{x - 1}\expo{-t}\,\dd t = 1 + \pars{x - 2}\int_{0}^{\infty}{t^{x - 1} - t \over x - 2}\expo{-t}\,\dd t \end{align}
For $\ds{x > 2\,,\ \exists\ \xi}$ such that $\ds{2 < \xi < x}$ and $\ds{{t^{x - 1} - t \over x - 2} = t^{\xi - 1}\ln\pars{t} \geq t\ln\pars{t}}$
\begin{align} \left.\vphantom{\Large A}\Gamma\pars{x}\,\right\vert_{\large\ x\ \geq\ 2}\ &\,\,\, {\large \geq}\,\,\, 1 + \pars{x - 2}\int_{0}^{\infty}t\ln\pars{t}\expo{-t}\,\dd t = 1 + \pars{x - 2}\pars{1 - \gamma}\ \bbox[10px,#ffe,border:1px dotted navy]{\ds{\geq 1}} \end{align}
By the Bohr-Mollerup characterization of the $\Gamma$ function, or from the Cauchy-Schwarz inequality applied to the integral representation, we have that $\Gamma(x)$ is a log-convex function on $\mathbb{R}^+$. Since $\Gamma(x)$ is positive for $x>0$, $\Gamma(x)$ is convex on $\mathbb{R}^+$ and
$$ \Gamma(x)\geq \Gamma'(3)(x-3)+\Gamma(3) $$ holds for any $x\in\mathbb{R}^+$. We have $\Gamma'(3)=3-2\gamma>1$, hence the given inequality is trivial.
In order to show that $0<\gamma<1$, it is enough to recall that $$ \gamma=\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right] $$ and that $x-\log(1+x)$ is bounded between $(1-\log(2))\,x^2$ and $\frac{1}{2}x^2$ on $(0,1)$.