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
$$f(r)=\int_{0}^{\infty}z^r d\mu(z)$$
is finite for all $r$ in the interval $I\subset \mathbb R$, 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
$$[\Gamma(s/n)]^n\leq\prod\limits_{1}^{n}\Gamma (x_i)\leq \left[\Gamma\left(\frac{s-m}{n-1}\right)\right]^{n-1}\Gamma(m).\qquad\qquad\qquad (1)$$

(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 $I\subset \mathbb R$. One may show that the function $\phi(x)=\prod\limits_{i=1}^{n}f(x_i)$, $x\in I^n$ is Schur-convex on $I^n$ if and only if $\log f$ is convex on $I$. Thus the function
$$\phi(x)=\prod\limits_{i=1}^n \Gamma(x_i),\quad x_i>0,\qquad \quad\qquad\qquad\qquad\qquad\qquad\quad (2)$$
is Schur-convex on $I^n=(0,\infty)^n$. Since $x_i\le m$, $i=1,\dots,n$, and $\sum x_i=s$, it is easy to check that
$$x \prec \left(\frac{s-m}{n-1},\dots,\frac{s-m}{n-1},m\right).$$
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 $x\succ (s/n,\dots,s/n)$.

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