Proving that $\frac{y}{2x(y + x)} - \log(x) + \log(y + x) + \psi(x) - \psi(y + x) \leq 0$ for $x \geq 1, y > 0$

Question

I am trying to evaluate a particular approximation to a distribution function. As a subproblem, I am trying to prove that for all $x \geq 1, y > 0$ we have:

$$ \frac{y}{2x(y + x)} - \log(x) + \log(y + x) + \psi(x) - \psi(y + x) \leq 0 $$

Where $\psi$ is the digamma function. This appears to hold because:

a) maximizing the left hand side numerically over $y,x$ yields a value within machine precision of $0$

b) Here's a plot for $1 < x < 100$ (horizontal axis), $0 < y < 10$ (vertical axis)

c) limit of the left hand size as $y \to 0$ as well as $x \to \infty$ is $0$

But I was not able to get very far in actually proving this is the case My hunch is that I am missing some bounds on the digamma function in terms of logarithms, but I could not find any in the NIST library.

The most related fact I was able to find is that $x(\log(x) - \psi(x)$ is decreasing in $x$ (Theorem 1.34 in https://journalofinequalitiesandapplications.springeropen.com/articles/10.1155/2010/493058), but I don't think that's enough to prove the full inequality.

Answer 1

Proof.

Let $$f(y) := \frac{y}{2x(y + x)} - \ln x + \ln(y + x) + \psi(x) - \psi(y + x).$$

We have $$f'(y) = \frac{1}{x + y} + \frac{1}{2(x + y)^2} - \psi'(y + x) \le 0$$ where we use $\psi'(u) > \frac{1}{u} + \frac{1}{2u^2}$ on $(0, \infty)$ (see the Remarks at the end).

Also, we have $f(0) = 0$. Thus, we have $f(y) \ge 0$ for all $y > 0$.

We are done.

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Remarks.

See (15) in this.

We can also use the fact that $\psi'(u) = \int_0^\infty \frac{t\mathrm{e}^{-ut}}{1 - \mathrm{e}^{-t}}\, \mathrm{d} t$ to prove it. Indeed, we have \begin{align*} \psi'(u) - \frac{1}{u} - \frac{1}{2u^2} &= \int_0^\infty \frac{t\mathrm{e}^{-ut}}{1 - \mathrm{e}^{-t}}\, \mathrm{d} t - \int_0^\infty \mathrm{e}^{-ut}\,\mathrm{d} t - \frac12 \int_0^\infty t \mathrm{e}^{-ut}\,\mathrm{d} t\\[6pt] &= \int_0^\infty \left(\frac{t\mathrm{e}^{-ut}}{1 - \mathrm{e}^{-t}} - \mathrm{e}^{-ut} - \frac12 t\mathrm{e}^{-ut}\right)\,\mathrm{d} t\\[6pt] &= \int_0^\infty \frac{(t + 2)\mathrm{e}^{-t} + t - 2}{2(1 - \mathrm{e}^{-t})}\cdot \mathrm{e}^{-ut}\,\mathrm{d} t\\[6pt] &> 0 \end{align*} where we use $(t + 2)\mathrm{e}^{-t} + t - 2 \ge 0$ on $[0, \infty)$ (the proof is easy).