# One particular probabilistic inequality

I've been studying for my exam in Probability theory and I found this exercise: prove that for every $\lambda$ greater than 0 this inequality holds: $$\int_{0}^{+\infty}\log(x+1)\lambda e^{-\lambda x}dx\geq \frac{1}{e}\log\left(\frac{\lambda +1}{\lambda}\right)$$ The left side is obviously equal to $\mathbb{E}[\log(X+1)]$, where $X\sim \text{Exp}(\lambda)$, but I don't see what right side might be. Can anyone provide me some help?

• Are you looking for some special meaning for this quantity ? – ippiki-ookami Jul 17 '18 at 14:45
• That is correct – Martin Jul 17 '18 at 14:48
• Is that the end of the exercise or are there follow up questions to that one ? – ippiki-ookami Jul 17 '18 at 14:49
• The whole exercise is to prove the inequality – Martin Jul 17 '18 at 14:50
• Please choose a more expressive title. The current title says basically nothing. – M. Winter Jul 17 '18 at 16:37

$\newcommand{\E}{\mathbb E}\newcommand{\PM}{\mathbb P}$Let $X\sim \text{Exp}(\lambda)$ then your problem is equivalent with proving the following: \begin{align} \E[\log(X+1)]\geq \PM(X>\E[X]) \log\E[X+1] \end{align} We know that \begin{align} \PM(\log(X+1)\geq \log\E[X+1] ) \leq \frac{\E[\log(X+1)]}{\log\E[X+1]} \end{align} By Markov's inequality, hence: \begin{align} \E[\log(X+1)]\geq \PM(\log(X+1)\geq \log\E[X+1] )\log\E[X+1] \end{align} But we also know: $$\PM(\log(X+1)\geq \log\E[X+1])=\PM(X+1>\E[X+1]) =\PM(X>\E[X])$$ (btw this manipulations is exactly what lead me to this answer) and this proves the inequality.
The only thing that we did is noticing that $e^{-1}=\PM(X>\E[X])$ and Markov's inequality.