According to http://functions.wolfram.com/06.05.29.0006.01, for every $x\geq 2$ it is
$$
\left( \frac{x}{e}\right)^{x-1} \leq \Gamma(x) \leq \left( \frac{x}{2}\right)^{x-1},
$$
where $\Gamma$ is the Gamma function. How can this be proven?
Update 1: According to Qi and Chen (see eq. (10)), the following holds:
$$
\ln \Gamma(x) = (x-1)(\ln x – 1) + \phi(x),
$$
where
$$
\phi(x) = \int_0^\infty \left(\frac{1}{t}-\frac{1}{e^{t}-1}\right) e^{-t} \frac{1-e^{-(x-1)t}}{t}\mathrm{d}t,
$$
thus, (for the left-hand side of the given inequality) it suffices to prove that $\phi(x)\geq 0$ for all $x\geq 2$ because it is equivalently written as $(x-1)(\ln x – 1)\leq \ln \Gamma(x)$.
Update 2: We take the derivative of $\phi$ which is:
$$
\phi'(x)=\int_{0}^{\infty}\left(\frac{1}{t}-\frac{1}{e^{t}-1}\right) e^{-t(1+x)}\mathrm{d}t
$$
and notice that the integrated function is positive for all $t>0$ and $x\geq 2$, therefore $\phi'(x)>0$ and $\phi$ is increasing with $\phi(2)=0.3069$ (with MATLAB, but I guess there must be some way to show analytically that $\phi(2)>0$). Then, we have proven that $\phi(x)\geq 0$ for all $x\geq 2$. Any better ideas?
Still we need to prove the right-hand side of the given inequality.
Best Answer
Here's a sketch of an alternative proof for the lower bound. First, note that $$ x\cdot\frac{x^{x-1}}{e^{x-1}} = \frac{(x+1)^x}{e^x} \cdot \frac{e}{\big(1+\frac1x\big)^x} \ge \frac{(x+1)^x}{e^x} $$ Since $\Gamma(x+1)=x\Gamma(x)$, this means that the lower bound for $x$ implies the lower bound for $x+1$; so it suffices to show the inequality for $x\in[2,3]$. Now, $(\ln\Gamma)'(2)=1-\gamma$ (see Polygamma function), and $\ln\Gamma(x)$ is convex, so $$ (1-\gamma)(x-2) \le \ln\Gamma(x) $$ Thus it suffices to prove that $$ \ln \Bigl(\Bigl(\frac xe\Bigr)^{x-1}\Bigr)\le (1-\gamma)(x-2) \qquad\text{for $x\in[2,3]$.} $$ And since the LHS here is also convex, it suffices to check this at $x=2$ and $x=3$.