Derivation of Chebyshev's Inequality in Harold Larson's book In section 5.5 page 277 of Harold Larson's Introduction to Probability Theory and Statistical Inference 3rd edition he develops Chebyshev's inequality as follows:

Assume $X$ is a random variable and $\epsilon$ is any positive constant. Define the random variable $Y_\epsilon$ by
$$
Y_\epsilon = 1 \text{ if } |X| \ge \epsilon
$$
$$
  \text{}\text{ }\text{ }\text{ }\text{ }         = 0 \text{ if } |X| \lt \epsilon
$$
Thus
$$
\mathbb E[Y_\epsilon] = 1 \cdot P(|X| \ge \epsilon) + 0 \cdot P(|X| \lt \epsilon) = P(|X| \ge \epsilon)
$$
and it certainly must be true that $X^2 \ge X^2Y_\epsilon \ge \epsilon^2 Y_\epsilon$
From this it also follows that
$$
\mathbb E[X^2] \ge \mathbb E[\epsilon^2 Y_\epsilon] = \epsilon^2 \mathbb E[Y_\epsilon] = \epsilon^2 P(|X| \ge \epsilon)
$$
and thus
$$
P(|X| \ge \epsilon) \le \frac{1}{\epsilon^2} \mathbb E[X^2]
$$
Which gives the Chebyshev inequality.

I'm struggling to understand how he goes from
$$X^2 \ge X^2Y_\epsilon$$
to the inequality
$$ X^2Y_\epsilon \ge \epsilon^2 Y_\epsilon$$
It doesn't seem obvious to me.

If $X^2 Y_\epsilon \geq \epsilon^2 Y_\epsilon$, then $X^2$ must be larger then $\epsilon^2$...
...if  $Y_\epsilon \neq 0$. And the latter is only the case when $|X| > \epsilon$.
