Question: 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
$$I\_m \, \frac{I\_{m+1} + I\_{m-1}}{2} \geq I\_m^2 \leq I\_{m+1} I\_{m-1} \, .$$

Answer: 
*Too long for a comment.*

After marty cohen's answer, I think that we can prove it that it holds even for $m=n$.
$$I\_m=\int\_0^\infty (1+s)^m \, {\rm e}^{-ns} \, ds=e^n E\_{-m}(n)=\frac{e^n}{n^{m+1}} \Gamma (m+1,n)\qquad \text{if}\qquad \Re(n)>0$$

Now, considering
$$f(m)=I\_{m} \left( I\_{m+1} + I\_{m-1} \right) - 2 \, I\_{m+1} I\_{m-1}$$
$$e^{-2 n} n^{2 m+3}f(m)=n \Gamma (m,n) (n \Gamma (m+1,n)-2 \Gamma (m+2,n))+\Gamma (m+1,n) \Gamma (m+2,n)\tag 1$$

If $m=n$ and simplifying,  

$$f(n)=\frac{2 n+1}{n^3}-\frac{e^{2 n} } {n^{2 n+1} }\, \Gamma (n,n)^2$$ which is always positive.
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 $m, n$.