Proof of complete monotonicity of a binomial function

The following is the completely monotonic claim that actually holds (also hinted by Iosif Pinelis).

Claim. Let $f(x)=\log\binom{x}{px}$; then, $f''$ is CM.

We prove this claim as a corollary of the following impressive generalization.

Theorem. (Karp and Prilepkina, 2015). Consider the ratio \begin{equation*} W(x) := \frac{\prod_{i=1}^p \Gamma(A_ix+a_i)}{\prod_{j=1}^q\Gamma(B_jx+b_j)}, \end{equation*} where $(A_i,a_i)$ and $(B_j,b_j)$ are positive scalars. Then, $(\log W(x))''$ is CM if and only if \begin{equation*} P(u) = \sum_{i=1}^p \frac{e^{-a_iu/A_i}}{1-e^{-u/A_i}} - \sum_{j=1}^q \frac{e^{-b_ju/B_j}}{1-e^{-u/B_j}} \ge 0,\quad\forall u > 0. \end{equation*} In the affirmative case, $(\log W(x))'' = \int_0^\infty e^{-xu}uP(u)du$.

Proof of the claim. Let $q:= 1-p$. We need to check positivity of \begin{equation*} k(u) := \frac{e^{-u}}{1-e^{-u}} - \frac{e^{-u/p}}{1-e^{-u/p}} - \frac{e^{-u/q}}{1-e^{-u/q}}. \end{equation*} Observe that for $p\to 0$ or $p\to 1$, $k(u)\to 0$, which is promising. Simplifying, we reduce $k\ge 0$ to showing positivity of \begin{equation*} g(u) := \frac{1}{e^u-1} - \frac{1}{e^{u/p}-1} - \frac{1}{e^{u/q}-1}, \end{equation*} which I leave to the OP to verify.

This is a not a complete answer, but possibly a first step in it.

Using the Dirichlet formula (Theorem~1.6.1 in G. E. ANDREWS, R. ASKEY, AND R. ROY, Special functions, volume 71 of Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 1999) and then the substitution $v:=\frac1{1+z}$, one has \begin{equation} \psi(x+h)-\psi(x)=\int_0^\infty\Big(\frac1{(1+z)^x}-\frac1{(1+z)^{x+h}}\Big)\frac{d z}z =\int_0^1 v^{x-1}\frac{1-v^h}{1-v}d v \end{equation} for positive $x$ and $h$. So, \begin{equation} f'(x)=\int_0^1(pv^{px}+qv^{qx}-v^x)\frac{dv}{1-v}, \end{equation} where $q:=1-p$. In particular, it follows that $f'$ is positive (rather than negative). However, it appears that $f''$ is indeed completely monotone.

This is a footnote to Suvrit's answer: He reduces the problem to verifying the condition $$\frac{1}{e^u-1} \geq \frac{1}{e^{u/p}-1} + \frac{1}{e^{u/q}-1}$$ for $u>0$, $p+q=1$ and $p,q > 0$. Put $g(p) = \tfrac{1}{e^{u/p}-1}$ and extend continuously to $g(0)=0$. It is sufficient to show that $g$ is convex on $[0,1]$.

We have $$\frac{d^2 g}{(dp)^2} = \frac{e^{u/p} u}{p^3 (e^{u/p}-1)^2} \left( \frac{u}{p} \frac{e^{u/p}+1}{e^{u/p}-1}-2 \right)$$ so we just need to show the quantity in parentheses is positive. Putting $v=u/p$, we need to show $\tfrac{v (e^v+1)}{e^v-1} \geq 2$. We can rewrite this as $\tfrac{v}{2} \geq \tfrac{e^{v/2}-e^{-v/2}}{e^{v/2}+e^{-v/2}}= \tanh \tfrac{v}{2}$. Since the derivative of $\tanh x$ is $\tfrac{1}{\cosh^2 x}<1$, we have $x \geq \tanh x$. $\square$