for poisson distribution, show that $P(0<X<2(\lambda +1)) \ge \frac \lambda{\lambda +1}$

Question

If $X$ is poisson distributed with mean $\lambda \ge 0$, (integer). Show that

$P(0<X<2(\lambda +1)) \ge \frac \lambda{\lambda +1}$

I applied markov inequality but I can't match the answer exactly. Is there some other way of doing this?

Answer 1

So it turned out I was doing some calculation mistakes before. As suggested in the question, I got the answer this way:

$$P(0<X<2(\lambda +1))=1-P(X=0)+P(X\ge 2\lambda +2)$$ $$\implies P(X=0)+P(X\ge 2\lambda +2)\le 1-\frac{\lambda}{1+\lambda}$$ $$\implies e^{-\lambda}+e^{-\lambda}\sum_{k=2\lambda +2}^{\infty}\frac{\lambda^{k}}{k!}\le \frac 1{\lambda +1 }$$ multiplying with $e^{\lambda}$ both sides and writing taylor series of $e^{\lambda}$ on the RHS gives the required answer. {also use $\lambda \ge 1$ as this is trivially satisfied at $\lambda=0$.}