Understanding the Gamma Function Is this valid?
If $x_1 > x_2 > x_3 > 0$ and $\Delta{t_1} = \Delta{t_2} + \Delta{t_3}$,
Does it follow that:
$$\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}\frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}$$
I am thinking that this follows based on the following observations which seem to be true:
$$\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_1-1 + \Delta{t_1})}{\Gamma(x_1-1)}$$
and:
$$\frac{\Gamma(x_1+\Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1 + \Delta{t_1} - \Delta{t_2})}\frac{\Gamma(x_1 + \Delta{t_1} - \Delta{t_2})}{\Gamma(x_1)}$$
So that we have:
$$\frac{\Gamma(x_1+\Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_1 + \Delta{t_2} + \Delta{t_3})}{\Gamma(x_1 + \Delta{t_3})}\frac{\Gamma(x_1 + \Delta{t_3})}{\Gamma(x_1)}$$
$$\frac{\Gamma(x_1 + \Delta{t_2} + \Delta{t_3})}{\Gamma(x_1 + \Delta{t_2})} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}$$
$$\frac{\Gamma(x_1 + \Delta{t_3})}{\Gamma(x1)} \ge \frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}$$
 A: The answer is yes (assuming that $\Delta t_1$, $\Delta t_2$, and $\Delta t_3$ are all nonnegative).
Define $\psi(u)$ to be the digamma function $\psi(u) = \frac d{du}\big( \log \Gamma(u) \big) = \Gamma'(u)/\Gamma(u)$; it is known that $\psi$ is increasing on $(0,\infty)$ (which is equivalent to $\Gamma$ being log-convex). Note that
$$
\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}\frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}
$$
is equivalent to
$$
\log{\Gamma(x_1 + \Delta{t_1})}-\log{\Gamma(x_1)} \ge \log{\Gamma(x_2 + \Delta{t_2})}-\log{\Gamma(x_2)}+\log{\Gamma(x_3 + \Delta{t_3})}-\log{\Gamma(x_3)},
$$
which in turn is the same as
$$
\int_{x_1}^{x_1+\Delta t_1} \psi(u)\,du \ge \int_{x_2}^{x_2+\Delta t_2} \psi(u)\,du + \int_{x_3}^{x_3+\Delta t_3} \psi(u)\,du.
$$
Since $\psi$ is increasing, and $x_2<x_1$ and $x_3<x_1\le x_1+\Delta t_2$, we have
$$
\int_{x_2}^{x_2+\Delta t_2} \psi(u)\,du \le \int_{x_1}^{x_1+\Delta t_2} \psi(u)\,du
$$
and
$$
\int_{x_3}^{x_3+\Delta t_3} \psi(u)\,du \le \int_{x_1+\Delta t_2}^{x_1+\Delta t_2+\Delta t_3} \psi(u)\,du = \int_{x_1+\Delta t_2}^{x_1+\Delta t_1} \psi(u)\,du.
$$
Adding these last two inequalities establishes the desired result.
