Importance of Log Convexity of the Gamma Function

First, let me mention that log convexity of a function is implied by an analytic property, which appears to be more natural than log convexity itself. Namely, if $mu$ is a Borel measure on $[0,infty)$ such that the $r$th moment
$$f(r)=int_{0}^{infty}z^r dmu(z)$$
is finite for all $r$ in the interval $Isubset mathbb R$, then $log f$ is convex on $I$.

Log convexity can be effectively used in derivation of various inequalities involving the gamma function (particularly, two-sided estimates of products of gamma functions). It is linked with the notion of Schur convexity which is itself used in many applications.

An appetizer. Let $m=max x_i$, $s=sum x_i$, $x_i > 0$, $i = 1,dots,n$, then
$$[Gamma(s/n)]^nleqprodlimits_{1}^{n}Gamma (x_i)leq left[Gammaleft(frac{s-m}{n-1}right)right]^{n-1}Gamma(m).qquadqquadqquad (1)$$

(1) is trivial, of course, when all $x_i$ and $s/n$ are integers, but in general the bounds do not hold without assuming log convexity.

Edit added: a sketch of the proof. Let $f$ be a continuous positive function defined on an interval $Isubset mathbb R$. One may show that the function $phi(x)=prodlimits_{i=1}^{n}f(x_i)$, $xin I^n$ is Schur-convex on $I^n$ if and only if $log f$ is convex on $I$. Thus the function
$$phi(x)=prodlimits_{i=1}^n Gamma(x_i),quad x_i>0,qquad quadqquadqquadqquadqquadqquadquad (2)$$
is Schur-convex on $I^n=(0,infty)^n$. Since $x_ile m$, $i=1,dots,n$, and $sum x_i=s$, it is easy to check that
$$x prec left(frac{s-m}{n-1},dots,frac{s-m}{n-1},mright).$$
The latter majorization and the fact that $phi(x)$ defined by (2) is Schur-convex imply the upper bound (1). The lower bound follows from the standard majorization $xsucc (s/n,dots,s/n)$.

---

Have a look at the recent short article by Marshall and Olkin concerning this and related inequalities for the gamma function.