How to prove that $3^{x^2+x} (x+1)^{-x} \Gamma (x+1)\ge 1$ for $x>0$? Let 
$$f(x)=3^{x^2+x} (x+1)^{-x} \Gamma (x+1).$$
Drawing a picture with any computer algebra system, 
it is obviously that $f(x) \ge 1$ on $[0,\infty)$.

But How can we prove this? If we take derivative, then we get
$$
\frac{\mathrm d}{\mathrm dx}\log(f(x))=-\frac{x}{x+1}+(2 x+1) \log (3)-\log (x+1)+\psi(x+1),
$$
where $\psi(\cdot)$ is the digamma function.
Drawing a picture again, we see that this is positive and increasing

But again, how can we prove this?

Okay, I have a proof now for $x \in (0,1)$.
We can expand $\log(f(x))$ by this formula to get
$$
\log(f(x))=
\underset{t=2}{\overset{\infty }{\sum }}\frac{(-x)^t ((t-1) \zeta (t)-t)}{(t-1) t}+x^2 (3 \log )+x (3 \log -\gamma ).
$$
Thus it suffices to show that is decreasing for $t \ge 2$.
$$
\left|\frac{((t-1) \zeta (t)-t)}{(t-1) t}\right|
$$
This can be proved using this paper.

Here's a proof for $x > 1$.
If $c > 1$,
since
$x! > \sqrt{2\pi x}(x/e)^x$
for $x > 1$,
if $x > 1 $ then
$\begin{array}\\
f(x)
&=c^{x^2+x} (x+1)^{-x} \Gamma (x+1)\\
&=c^{x^2+x} (x+1)^{-x} x!\\
&>c^{x^2+x} (x+1)^{-x} \sqrt{2\pi x}(x/e)^x\\
&=\sqrt{2\pi x}\left(c^{x+1} \dfrac{x}{e(x+1)}\right)^x\\
&>\sqrt{2\pi x}\left( \dfrac{c^2x}{e(x+1)}\right)^x\\
&>\sqrt{2\pi x}\left( \dfrac{c^2}{2e}\right)^x
\qquad\text{since } x/(x+1) > 1/2 \text{ for } x > 1\\
\end{array}
$
Therefore,
if $c^2 > 2e$,
or $c > 2.34 > \sqrt{2e}
$,
$f(x) \gt \sqrt{2\pi x}$
for $x > 1$.
