[proofplan]
The event $\{X = 0\}$ forces the deviation $|X - \mathbb{E}[X]|$ to equal $\mathbb{E}[X]$, since $X$ is a non-negative integer. Hence $\{X = 0\}$ is contained in the tail event $\{|X - \mathbb{E}[X]| \geq \mathbb{E}[X]\}$, to which [Chebyshev's inequality](/theorems/???) applies with threshold $t = \mathbb{E}[X]$. The Chebyshev bound delivers the stated ratio $\operatorname{Var}(X)/(\mathbb{E}[X])^2$.
[/proofplan]
[step:Dispose of the degenerate case $\mathbb{E}[X] = 0$]
If $\mathbb{E}[X] = 0$, then since $X \geq 0$ takes values in $\mathbb{N} = \{0, 1, 2, \ldots\}$, we have $X = 0$ almost surely, so $\mathbb{P}(X = 0) = 1$ and $\operatorname{Var}(X) = 0$. The inequality reads $1 \leq 0/0$, which is interpreted conventionally as vacuous (the right-hand side is the indeterminate form and the statement is understood to hold vacuously, or equivalently we exclude this case). For the remainder of the proof we assume $\mathbb{E}[X] > 0$, which is the only case where the ratio on the right-hand side is well-defined.
[/step]
[step:Translate $\{X = 0\}$ into a deviation event]
Since $X$ takes values in $\mathbb{N}$, the event $\{X = 0\}$ is a pointwise condition. On the event $\{X = 0\}$ we compute
\begin{align*}
|X - \mathbb{E}[X]| = |0 - \mathbb{E}[X]| = \mathbb{E}[X],
\end{align*}
where we used $\mathbb{E}[X] \geq 0$ to drop the absolute value. In particular, the pointwise inequality $|X - \mathbb{E}[X]| \geq \mathbb{E}[X]$ holds on $\{X = 0\}$, giving the set-theoretic inclusion
\begin{align*}
\{X = 0\} \subseteq \{|X - \mathbb{E}[X]| \geq \mathbb{E}[X]\}.
\end{align*}
By monotonicity of probability,
\begin{align*}
\mathbb{P}(X = 0) \leq \mathbb{P}(|X - \mathbb{E}[X]| \geq \mathbb{E}[X]).
\end{align*}
[guided]
The key observation is that if $X$ is supported on $\mathbb{N}$ and vanishes, then $X$ is as far as possible from its mean in a specific quantitative sense: the deviation is exactly $\mathbb{E}[X]$. This is a peculiarity of non-negative integer random variables — it would fail for, say, centred Gaussians, where $X = 0$ is no deviation at all.
Concretely, on the event $\{X = 0\}$ we have the pointwise identity
\begin{align*}
|X - \mathbb{E}[X]| = |0 - \mathbb{E}[X]| = \mathbb{E}[X]
\end{align*}
(using $\mathbb{E}[X] \geq 0$ since $X \geq 0$, so the absolute value is superfluous). Therefore on $\{X = 0\}$ the inequality $|X - \mathbb{E}[X]| \geq \mathbb{E}[X]$ is satisfied — in fact with equality. This gives the event inclusion
\begin{align*}
\{X = 0\} \subseteq \{|X - \mathbb{E}[X]| \geq \mathbb{E}[X]\},
\end{align*}
and monotonicity of the probability measure yields
\begin{align*}
\mathbb{P}(X = 0) \leq \mathbb{P}(|X - \mathbb{E}[X]| \geq \mathbb{E}[X]).
\end{align*}
We have converted a question about a point in the support of $X$ into a tail-deviation question, which is exactly the form that Chebyshev's inequality controls.
[/guided]
[/step]
[step:Apply Chebyshev's inequality with threshold $t = \mathbb{E}[X]$]
We invoke [Chebyshev's inequality](/theorems/???): for any random variable $Y$ with finite second moment and any $t > 0$,
\begin{align*}
\mathbb{P}(|Y - \mathbb{E}[Y]| \geq t) \leq \frac{\operatorname{Var}(Y)}{t^2}.
\end{align*}
The hypotheses are met with $Y = X$ and $t = \mathbb{E}[X] > 0$: we have $t > 0$ by the assumption of the previous step, and if $\operatorname{Var}(X) = \infty$ the conclusion is vacuous. Applying the inequality,
\begin{align*}
\mathbb{P}(|X - \mathbb{E}[X]| \geq \mathbb{E}[X]) \leq \frac{\operatorname{Var}(X)}{(\mathbb{E}[X])^2}.
\end{align*}
Combining with the inclusion from the previous step,
\begin{align*}
\mathbb{P}(X = 0) \leq \mathbb{P}(|X - \mathbb{E}[X]| \geq \mathbb{E}[X]) \leq \frac{\operatorname{Var}(X)}{(\mathbb{E}[X])^2},
\end{align*}
which is the desired inequality.
[guided]
We now call on [Chebyshev's inequality](/theorems/???), which states: for any random variable $Y$ with $\mathbb{E}[Y^2] < \infty$ and any threshold $t > 0$,
\begin{align*}
\mathbb{P}(|Y - \mathbb{E}[Y]| \geq t) \leq \frac{\operatorname{Var}(Y)}{t^2}.
\end{align*}
We verify the hypotheses for the application $Y := X$, $t := \mathbb{E}[X]$:
(i) the threshold $t = \mathbb{E}[X]$ is strictly positive by the degenerate-case reduction in the first step;
(ii) if $\operatorname{Var}(X) = \infty$ the right-hand side of the claimed inequality is $\infty$ and there is nothing to prove, so we may assume $\operatorname{Var}(X) < \infty$, which together with $\mathbb{E}[X] < \infty$ gives $\mathbb{E}[X^2] = \operatorname{Var}(X) + (\mathbb{E}[X])^2 < \infty$.
Applying Chebyshev with these choices yields
\begin{align*}
\mathbb{P}(|X - \mathbb{E}[X]| \geq \mathbb{E}[X]) \leq \frac{\operatorname{Var}(X)}{(\mathbb{E}[X])^2}.
\end{align*}
Chaining with the inclusion $\{X = 0\} \subseteq \{|X - \mathbb{E}[X]| \geq \mathbb{E}[X]\}$ established in the previous step,
\begin{align*}
\mathbb{P}(X = 0) \leq \mathbb{P}(|X - \mathbb{E}[X]| \geq \mathbb{E}[X]) \leq \frac{\operatorname{Var}(X)}{(\mathbb{E}[X])^2}.
\end{align*}
This is the content of the theorem, often called the *second moment method*: a small variance-to-mean-squared ratio forces $X$ to be positive with high probability. The integrality of $X$ is essential — the bound encodes the fact that integer-valued random variables can only reach zero by deviating from the mean by the full amount $\mathbb{E}[X]$.
[/guided]
[/step]
