If $X$ is Poisson distributed with parameter $n$ then $P(0<X<2(n+1))\ge \frac{n}{n+1}$

Question

If $X$ is Poisson distributed with parameter $n$ then $P(0<X<2(n+1))\ge \frac{n}{n+1}$.

I have tried to approach the problem but I just cannot figure out how to do it. I tried expanding the $e^{-x}$ but it just brought out a lo of terms which I don't know how to handle. Please help me out.

Edit: Ok guys I have tried the following using spaceisdarkGreen's suggestion. Now here is the part I have tried.

$$P(X>= 2(n+1) <= \frac{E(X)}{2(n+1)}$$ $$ E(X) = \int{\frac{e^{-x}x^{n+1}}{n!}}dx$$ $$=n+1$$ Therefore,

$$P(X>=2(n+1))<=\frac{1}{2}$$ $$ 1-P(0<X<2(n+1))<=\frac{1}{2}$$ $$ P(0<X<2(n+1))>=\frac{1}{2}$$

Answer 1

Use Chebyshev's inequality, often stated as $$ P(|X-E(X)| \ge a) \le \frac{Var(X)}{a^2}$$ for any random variable with finite expectation.

As a further hint, you can write $$P(X \ge 2(n+1)) = P((X-(n+1))\ge n+1) \le P(|X-(n+1)| \ge n+1).$$