# 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 $$x > 1$$.
• This is correct. Though I actually care more about $0<x<1$. – ablmf Oct 22 '18 at 15:35