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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 $[0,\infty)$.

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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

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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.

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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 $x > 1$.

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

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