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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.

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    Thank you for your comment, I've edited the question. – 2017-02-09
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    Just something I noticed: $$\int_0^\infty\frac{x^{s-1}}{e^x+z}\ dx=\frac1{-z}\text{Li}_s(-z)\Gamma(s)$$where I used [Polylogarithms](https://en.m.wikipedia.org/wiki/Polylogarithm) – 2017-02-09
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    use integration by parts for the integral on the rhs of your inequality – 2017-02-09
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    Furthermore $(e^x+z)^2>e^x+z$ for $(x,z)\in\mathbb{R}^2_{+}$ – 2017-02-09
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    @SimplyBeautifulArt note that this integral is effectively a shifted form of the [Fermi Dirac Integral](https://en.m.wikipedia.org/wiki/Complete_Fermi%E2%80%93Dirac_integral) which quickly confirms your work with the Polylogarithm – 2017-02-09
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    @SimplyBeautifulArt moreover, your inequality is confirmed [here](http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/29/ShowAll.html) – 2017-02-09
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    @SimplyBeautifulArt I'll post an answer elaborating on this when I find time – 2017-02-09
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    @BrevanEllefsen No need, I already finished it up. – 2017-02-09

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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:

$$\frac d{ds}\text{Li}_s(-z)=-\sum_{n=1}^\infty\frac{(-z)^n\ln(n)}{n^s}<-\sum_{n=1}^2\frac{(-z)^n\ln(n)}{n^s}<0$$

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    Good answer. I still think the connection to the Fermi-Dirac integral is worth mentioning though... it is always nice to know when something is in the literature IMO :) – 2017-02-09
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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.