When is log-convexity inequality for Gamma function tight?

113 Views Asked by At

Is the following is a well known inequality for the Gamma function. If $\alpha \in [0,1]$ then \begin{align} \Gamma \left( (1-\alpha)x+\alpha y \right) \le \Gamma^\alpha \left( x \right)\Gamma^{1-\alpha} \left( y \right) \end{align}

For which instance is this inequality tight (except trivial once $\alpha=0,1$)? I tried a few example the were almost there but not exactly.

1

There are 1 best solutions below

1
On BEST ANSWER

The inequality is tight only for the trivial cases $\alpha=0,1$, because the function $\log\Gamma(x)$ is strictly convex in the interval $(0,\infty)$. This means that for any $x<y$, the graph of the chord connecting the points $(x,\log\Gamma(x))$ and $(y,\log\Gamma(y))$ will be strictly above the graph, touching it only at those two points, which correspond to $\alpha=0,1$.