The minimum of the reciprocals of some Poisson random variables Let $X_1,\dots,X_k$ denote a collection of independent samples of a Poisson random variable whose mean also happens to be equal to $k$.  Does the quantity $$k\boldsymbol{E}\min\left\{ \frac{1}{1+X_{1}},\dots,\frac{1}{1+X_{k}}\right\} $$ have a strictly positive limit as $k$ becomes large?
 A: Let 
\begin{equation}
 Y_i:=\frac1{1+X_i}, \quad Y:=\min(Y_1,\dots,Y_k). 
\end{equation}
Then 
\begin{equation}
 EY=\int_0^\infty P(Y>y)\,dy=\int_0^\infty P(Y_1>y)^k\,dy. 
\end{equation}
Next, for $y\in(0,\frac1{1+2k})$ and $x:=\frac1y-1>2k$, we have 
\begin{equation}
 P(Y_1>y)=1-P(X_1>x),\quad P(X_1>x)\le P(X_1>2k)=P(X_1-k>k)\le1/k,
\end{equation}
by Chebyshev's inequality, whence $P(Y_1>y)\ge1-1/k$ and 
\begin{equation}
 EY\ge\int_0^{1/(1+2k)} P(Y_1>y)^k\,dy\ge\int_0^{1/(1+2k)} (1-1/k)^k\,dy\sim\frac1{2ek}
\end{equation}
as $k\to\infty$. So, if the limit of $k\,EY$ exists, it must be $\ge\frac1{2e}>0$. 
A: (To answer Losif Pinelis, this was a bit too long for a comment) 
For $X$ a Poisson of mean $k$, $$\mathbb{P}(X\geq (1+\epsilon)k) =\mathbb{P}(e^{X\epsilon/4}\geq e^{(1+\epsilon)k\epsilon/4}) \\ \leq \frac{\mathbb{E}(e^{X\epsilon / 4})}{e^{(1+\epsilon)k\epsilon/4}} =\frac{\exp(k(e^{\epsilon /4}-1))}{e^{(1+\epsilon)k\epsilon/4}} \\ \approx\exp(-k \epsilon^2(\frac{1}{4}-\frac{1}{2*4^2}) $$ for small $\epsilon$. One can then finish the proof of Pinelis and get that the limit is larger than $1/(1+\epsilon)$. 
