How to proof $\frac{I_{m+1} I_{m} + I_m I_{m-1}}{2} \geq I_{m+1} I_{m-1}$

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 1

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 $m \ge 2n$.

Answer 2

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 $m, n$.

Answer 3

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.