<s>[INST] The exact constant in a bound on ratios of Gamma functions The answer to another question (Upper bound of the fraction of Gamma functions) gave an asymptotic upper bound for an expression with Gamma functions:
$$\left(\frac{\Gamma(a+b)}{a\Gamma(a)\Gamma(b)}\right)^{1/a}\!\leq \,C\,\frac{a+b}a, \forall a,b\geq\frac12$$
What is the best possible value for the constant $C$ in that statement?
 [/INST] The optimal $C$ is $\mathrm{e}$.
Proof:
We have
$$\ln C \ge \ln a - \ln(a + b)
+ \frac{\ln \Gamma(a + b)
    -\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$
Let
$$F(a, b) := \ln a - \ln(a + b)
+ \frac{\ln \Gamma(a + b)
    -\ln a - \ln\Gamma(a) - \ln\Gamma(b)}{a}.$$
We have
$$\frac{\partial F}{\partial b}
= - \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a} \ge 0 \tag{1}$$
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$.
The proof of (1) is given at the end.
Fixed $a\ge 1/2$, we have
$$G(a) := \lim_{b\to \infty} F(a, b) = 
\ln a 
+ \frac{
    -\ln a - \ln\Gamma(a)}{a}
$$
where we have used
$$\lim_{b\to \infty} -\ln(a + b) + \frac{\ln \Gamma(a + b) - \ln\Gamma(b)}{a} = 0.$$
(Note: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.)
We have
$$
    a^2 G'(a) = a - 1 - a\psi(a) + \ln a + \ln \Gamma(a) \ge 0.\tag{2}
$$
The proof of (2) is given at the end.
We have
$$\lim_{a\to \infty} G(a) = 1.$$
(Note: Use $\sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x} \le \Gamma(x) \le \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}\mathrm{e}^{\frac{1}{12x}}$ for all $x > 0$.)
Thus, the optimal $C$ is $\mathrm{e}$.

Proof of (1):
Using Theorem 5 in [1]: for all $u > 0$,
$$\ln u - \frac{1}{2u} - \frac{1}{12u^2} < \psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2},$$
we have
\begin{align*}
    &- \frac{1}{a+b} + \frac{\psi(a + b) - \psi(b)}{a}\\
    \ge{}&- \frac{1}{a+b} + \frac{1}{a} \left(\ln (a+b) - \frac{1}{2(a+b)} - \frac{1}{12(a+b)^2}\right)\\
    &\qquad
    - \frac{1}{a}\left(\ln b - \frac{1}{2b} - \frac{1}{12(b+1/14)^2}\right)\\
    ={}& \frac{1}{a}\ln(1 + a/b) - \frac{1}{a+b} - \frac{1}{2a(a+b)} - \frac{1}{12a(a+b)^2} + \frac{1}{2ab} + \frac{1}{12a(b+1/14)^2}\\
    \ge{}& \frac{1}{a}\left(\ln(1 + a/b) - \frac{a/b}{1 + a/b}\right)
    + \frac{1}{2ab} - \frac{1}{2a(a+b)} + \frac{1}{12a(b+1/14)^2} - \frac{1}{12a(a+b)^2}\\
    \ge{}&0
\end{align*}
where we use $\ln(1+x) \ge \frac{x}{1+x}$ for all $x \ge 0$.
We are done.
$\phantom{2}$
Proof of (2):
Using
$\Gamma(x) \ge \sqrt{2\pi}\, x^{x-1/2}\mathrm{e}^{-x}$ and
$\psi(u) < \ln u - \frac{1}{2u} - \frac{1}{12(u+1/14)^2}$ for all $u > 0$ (Theorem 5 in [1]),
we have
\begin{align*}
    &a - 1 - a\psi(a) + \ln a + \ln \Gamma(a)\\
    \ge{}& a - 1 - a \left(\ln a - \frac{1}{2a} - \frac{1}{12(a+1/14)^2}\right) + \ln a + \frac12\ln(2\pi) + (a-1/2)\ln a - a\\
    ={}& \frac12\ln(2\pi a) - \frac12 + \frac{a}{12(a+1/14)^2}\\
    \ge{}& 0.
\end{align*}
We are done.

Reference
[1] L. Gordon, “A stochastic approach to the gamma function”, Amer. Math. Monthly, 9(101), 1994, 858-865.
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