= Solution
Because $X$ is a nonnegative integer, its <indicator function> satisfies $\mathbf1_{\{X>0\}}\leq X$. Taking <expected values> proves the upper bound. The finite <second moment> also gives a finite <expected value>, by the <Cauchy-Schwarz inequality>. Applying that inequality to $X$ and $\mathbf1_{\{X>0\}}$ gives
$$
(\mathbb EX)^2
=\bigl(\mathbb E[X\mathbf1_{\{X>0\}}]\bigr)^2
\leq\mathbb E[X^2]\,\mathbb P(X>0).
$$
The assumed positive <second moment> permits division, yielding
$$
\boxed{\frac{(\mathbb EX)^2}{\mathbb E[X^2]}\leq\mathbb P(X>0)\leq\mathbb EX.}
$$
The lower bound is the basic <second moment method>; the upper bound also follows from <Markov's inequality> at threshold one. Integer-valuedness is essential for that upper bound, but not for the lower bound. For example, a constant $X=1/2$ would violate the upper bound.
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