We are using the general definition of gamma function defined on $\Bbb C \setminus \{\text{non-positive integers}\}$ i.e. $\Gamma(Z)=\frac {e^{-\gamma z}}{z} \prod (1+ \frac zn)^{-1} e^{\frac zn}$. Note that $\Gamma(x) \gt 0$ on $x \gt 0$.
To show that $\Gamma$ is logarithmically convex on $(0,\infty)$, we first need to show that it is twice differentiable on $x \gt 0$.
Query : Can we say that $\Gamma$ is twice differentiable on $(0,\infty)$ because it is analytic on right half plane? If so then how to prove that? If not so then what is the way?
If we show this then we can differentiate $\log \Gamma(x)$ w.r.t $x$ to get $\frac {d}{dx} \log \Gamma(x)=\frac {\Gamma'(x)}{\Gamma(x)}$ and then by observing $\frac {d^2}{dx^2} \log \Gamma(x) = \left(\frac {\Gamma'(x)}{\Gamma(x)}\right)' = \frac 1{x^2} + \sum_{n=1}^{\infty} \frac {1}{(n+x)^2} \gt 0$, we can conclude that $\Gamma$ is logarithmically convex on $x \gt 0.$
Show the infinite product $$\prod_{n=1}^\infty (1-n^{-1}z)^{-1}\exp(z/n)$$ converges uniformly on compact subsets away from the poles $-1,-2,\dots$. Hence $\Gamma$, being local uniform limit of holomorphic functions on $\mathbb{C}-\{0,-1,-2,\dots\}$, is holomorphic there by Morrera.