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.
I will show that
the inequality
is true for
$m$ large compared to $n$.

$\begin{array}\\
I\_m
&=\int\_0^\infty (1+s)^m e^{-ns}ds\\
&=\int\_0^\infty \sum\_{k=0}^m{m\choose k}s^k e^{-ns}ds\\
&=\sum\_{k=0}^m{m \choose k}\int\_0^\infty s^k e^{-ns}ds\\
&=\sum\_{k=0}^m{m \choose k}\frac{k!}{n^{k+1}}\\
&=\sum\_{k=0}^m\frac{m!}{(m-k)!n^{k+1}}\\
&=\sum\_{k=0}^m\frac{m!}{k!n^{m-k+1}}\\
&=\frac{m!}{n^{m+1}}\sum\_{k=0}^m\frac{n^k}{k!}\\
&<\frac{m!}{n^{m+1}}e^n
\\
\end{array}
$

If we treat this
as an equality,
and it is close
for $m >> n$,
then
$\frac1{I\_m}
\approx \frac{n^{m+1}}{m!e^n}
$
so that
$\frac{1}{I\_{m+1}} + \frac{1}{I\_{m-1}} \geq \frac{2}{I\_m}
$
becomes
$ \frac{n^{m+2}}{(m+1)!e^{n}}+ \frac{n^{m}}{(m-1)!e^n} \geq \frac{2n^{m+1}}{m!e^n}
$

Dividing by
$\frac{n^m}{(m-1)!e^n}
$
this becomes
$ \frac{n^2}{m(m+1)}+ 1 \geq \frac{2n}{m}
$
which is certainly true
for $m \ge 2n$.