For $s \geq 0$ and $x > 0$ define $$ f_s (x) = - \operatorname{Li}_s (-x) \stackrel{s > 0}{=} \frac{1}{\Gamma(s)} \int \limits_0^\infty \frac{x t^{s-1}}{\mathrm{e}^t + x} \, \mathrm{d} t \, .$$ I would like to prove the inequality $$ \tag{1} (s+1) \frac{f_{s+1}(x)}{f_s(x)} > \log(x) $$ for $s \geq 0$ and $x \geq 1$ (it is obviously correct for $x \in (0,1)$). It can be used to show that the entropy of an ideal Fermi gas is always non-negative. Here's what I have come up with so far:
Since $f_s (1) = \eta (s) > 0$ for $s \geq 0$ ($\eta$ is the Dirichlet eta function), $(1)$ trivially holds for $x = 1$. Note that we have $x f_{s+1}'(x) = f_s (x)$ for $s \geq 0$ and $x > 0$ by definition. Now suppose that $(1)$ is true for some $s \geq 0$. Then $$ \frac{\mathrm{d}}{\mathrm{d}x} [(s+2)f_{s+2}(x) - \log(x) f_{s+1}(x)] = \frac{1}{x} [(s+1)f_{s+1}(x) - \log(x) f_{s}(x)] > 0$$ follows for $x \geq 1$ and integration yields $(1)$ with $s$ replaced by $s+1$. Therefore, it is sufficient to prove the inequality for $s \in [0,1)$.
The case $s=0$ is trivial, as $f_0 (x) = \frac{x}{1+x}$ and $f_1(x) = \log(1+x)$. Now let $s \in (0,1)$. The inequality $f_{s+1}(x) > f_{s}(x) , \, x > 0 ,$ confirms $(1)$ for $x \in [1,\mathrm{e}^{s+1}]$. The asymptotic expansion $$ f_{s} (x) \stackrel{x \to \infty}{\sim} 2 \log^s (x) \sum \limits_{k=0}^\infty \frac{\eta(2k)}{\Gamma(s+1-2k) \log^{2k}(x)} $$ leads to $$ \frac{(s+1) f_{s+1}(x)}{\log(x) f_s(x)} \stackrel{x \to \infty}{\sim} 1 + \frac{\pi^2 s}{3 \log^2 (x)} + \mathcal{O}(\log^{-4}(x)) \, , $$ which shows that $(1)$ holds for sufficiently large $x$. These results and numerical experiments clearly suggest that the inequality is true for all $x \geq 1$, but I have not found a proof yet.
Integration by parts reveals that $(1)$ is equivalent to $$ \tag{2} \int \limits_0^\infty \frac{u^{s+1} \mathrm{e}^{\alpha (u-1)}}{(\mathrm{e}^{\alpha (u-1)} + 1)^2} \, \mathrm{d} u > \int \limits_0^\infty \frac{u^s \mathrm{e}^{\alpha (u-1)}}{(\mathrm{e}^{\alpha (u-1)} + 1)^2} \, \mathrm{d} u \, , \, \alpha = \log(x),$$ but I am stuck here as well, so my question is:
How can we prove $(1)$ or $(2)$ for $s \in (0,1)$ and $x > 1$ ?
I have found a simple method to prove the inequality: Fix $x > 1$ and let $\alpha = \log(x) > 0$. Define $$ g \colon [0,\infty) \to \mathbb{R} \, , \, g(s) = \int \limits_0^\infty \frac{(u-1) u^s \mathrm{e}^{\alpha(u-1)}}{(\mathrm{e}^{\alpha (u-1)} + 1)^2} \, \mathrm{d} u \, . $$ Then $(1)$ is equivalent to $g(s) > 0$ for $s \geq 0$ (for $s=0$ this follows by analytic continuation). But $(1)$ holds for $s=0$, so $g(0) > 0$. Moreover, $g$ is differentiable with $$ g'(s) = \int \limits_0^\infty \frac{\log(u) (u-1) u^s \mathrm{e}^{\alpha(u-1)}}{(\mathrm{e}^{\alpha (u-1)} + 1)^2} \, \mathrm{d} u > 0 \, , \, s \geq 0.$$ This implies $g(s) > 0$, and hence $(1)$ for all $s \geq 0$.