Proving the inequality $\int\limits_{0}^{\infty} \frac{x^{s-1}}{e^{x}+z} dx \geq (s-1) \int\limits_{0}^{\infty} \frac{x^{s-2}}{e^{x}+z} dx$ I'm trying to prove this inequality 

$$\int\limits_{0}^{\infty} \frac{x^{s-1}}{e^{x}+z} dx \geq (s-1) \int\limits_{0}^{\infty} \frac{x^{s-2}}{e^{x}+z} dx$$

where $s>2$ and $z\in \mathbb{R}^{+}$. I checked with MATLAB for different values of $s,z$ and the results agree with the given inequality.
 A: Due to the geometric series and $(n+1)x=u$,
$$\begin{align}\int_0^\infty\frac{x^{s-1}}{e^x+z}\ dx&=\int_0^\infty x^{s-1}e^{-x}\frac1{1+ze^{-x}}\ dx\\
&=\int_0^\infty x^{s-1}e^{-x}\sum_{n=0}^\infty(-ze^{-x})^n\\
&=\sum_{n=0}^\infty(-z)^n\int_0^\infty x^{s-1}e^{-(n+1)x}\ dx\\
&=\sum_{n=0}^\infty\frac{(-z)^n}{(n+1)^s}\int_0^\infty u^{s-1}e^{-u}\ du\\
&=\frac1{-z}\text{Li}_s(-z)\Gamma(s)\end{align}$$
It follows that your inequality reduces to proving
$$\text{Li}_s(-z)<\text{Li}_{s-1}(-z)$$
This can be generalized into a more general statement for $a,z\in\mathbb R^+$:
$$\text{Li}_s(-z)<\text{Li}_{s-a}(-z)$$
Which is equivalent to proving
$$\frac d{ds}\text{Li}_s(-z)<0$$
which follows from the bounds of an alternating series:
$$\frac d{ds}\text{Li}_s(-z)=-\sum_{n=1}^\infty\frac{(-z)^n\ln(n)}{n^s}<-\sum_{n=1}^2\frac{(-z)^n\ln(n)}{n^s}<0$$
A: Define $f=x^{s-1},\,g=fe^{-x},\,h=\dfrac{e^x}{e^x+z}$. Subtracting the right-hand side of the inequality from the left gives $$\int_0^\infty \left(f-f'\right)e^{-x}hdx=-\int_0^\infty g'hdx=\int_0^\infty gh'dx$$(you can verify the boundary term of integration by parts vanishes). Since$$h'=\partial_x\frac{1}{1+ze^{-x}}=\frac{ze^{-x}}{\left( 1+ze^{-x}\right)^2}\ge 0,$$our original integral is non-negative as required.
