<s>[INST] How to proof $\frac{I_{m+1} I_{m} + I_m I_{m-1}}{2} \geq I_{m+1} I_{m-1}$ Let
$$ I_m=\int_0^\infty (1+s)^m \, {\rm e}^{-ns} \, {\rm d}s \, .$$
I've been thinking some time on the following inequality
$$I_{m} \left( I_{m+1} + I_{m-1} \right) \geq 2 \, I_{m+1} I_{m-1}$$
or in somewhat interesting parallel circuit fashion
$$\frac{1}{I_{m+1}} + \frac{1}{I_{m-1}} \geq \frac{2}{I_m} \, .$$
Is there any clever way to tackle it?
It also looks somewhat like AMGM
$$I_{m+1}^2 + I_{m-1}^2 \geq 2 \, I_{m+1}I_{m-1}$$
in a sense, where one $I_{m+1}$ and $I_{m-1}$ on the LHS is replaced by $I_m$.
$I_m$ is monotonically increasing in $m$ and log-convex (i.e. $I_{m+1} I_{m-1} \geq I_m^2$, without proof though).
It seems to be a rather tight inequality, since e.g. 
$$I_{m+1} I_m + I_m I_{m-1} \geq I_{m+\frac{1}{2}}^2 + I_{m-\frac{1}{2}}^2 \geq 2 \, I_{m+\frac{1}{2}} I_{m-\frac{1}{2}} \leq 2 \, I_{m+1} I_{m-1}$$
or
$$I_m \, \frac{I_{m+1} + I_{m-1}}{2} \geq I_m^2 \leq I_{m+1} I_{m-1} \, .$$
 [/INST] Expanding on marty cohen's answer, notice the recurrence
\begin{align*}
I_{m+1} = \frac{1}{n} + \frac{m+1}{n}I_m
\end{align*}
and
\begin{align*}
I_{m-1} = - \frac{1}{m} +\frac{n}{m}I_m
\end{align*}
Substituting this into the desired inequality,
\begin{align*}
I_m\left(\frac{1}{n} + \frac{m+1}{n}I_m - \frac{1}{m} + \frac{n}{m}I_m \right) \ge 2\left(\frac{1}{n} + \frac{m+1}{n}I_m\right)\left(- \frac{1}{m} + \frac{n}{m}I_m\right)
\end{align*}
or
\begin{align*}
I_m \ge \frac{\sqrt{(m - n - 2)^2 + 8m} + 3(n - m) - 2}{2[(m+1-n)^2 - (m+1)]}
\end{align*}
From here, you might try induction on $m, n$.
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