How to prove that $P(X = x-1) \cdot P(X=x+1) \le (P(X=x))^2$ for a Poisson distribution

Question

Using Poisson distribution, i need to prove that

$$P(X = x-1)\cdot P(X=x+1) \le (P(X=x))^2$$

How do i prove that this is true?

Answer 1

Hint:

Let $X\sim Poisson(\lambda)$. Then

$P(X=x-1)P(X=x+1)=\frac{\lambda^{x-1}e^{-\lambda}}{(x-1)!} \frac{\lambda^{x+1}e^{-\lambda}}{(x+1)!}=\frac{\lambda^{2x}e^{-2\lambda}}{(x-1)!(x+1)!}$.