How to understand the proof of the division theorem in the four arithmetic operation theorems of probability convergence for random variables?

Question

\begin{equation} Y_n \stackrel{P}{\longrightarrow} b \text {} \end{equation} and: \begin{aligned} & P\left(\left|\frac{1}{Y_n}-\frac{1}{b}\right| \geqslant \varepsilon\right)=P\left(\left|\frac{Y_n-b}{Y_n b}\right| \geqslant \varepsilon\right) \\ = & P\left(\left|\frac{Y_n-b}{b^2+b\left(Y_n-b\right)}\right| \geqslant \varepsilon,\left|Y_n-b\right|<\varepsilon\right)+P\left(\left|\frac{Y_n-b}{b^2+b\left(Y_n-b\right)}\right| \geqslant \varepsilon,\left|Y_n-b\right| \geqslant \varepsilon\right) \\ \leqslant & P\left(\frac{\left|Y_n-b\right|}{b^2-\varepsilon|b|} \geqslant \varepsilon\right)+P\left(\left|Y_n-b\right| \geqslant \varepsilon\right)=P\left(\left|Y_n-b\right| \geqslant\left(b^2-\varepsilon|b|\right) \varepsilon\right)+P\left(\left|Y_n-b\right| \geqslant \varepsilon\right) \rightarrow 0(n \rightarrow \infty) . \end{aligned} so \begin{equation} 1 / Y_n \stackrel{P}{\longrightarrow} 1 / b \end{equation} my question is how to understand this step: $$ P\left(\left|\frac{Y_n-b}{b^2+b\left(Y_n-b\right)}\right| \geqslant \varepsilon,\left|Y_n-b\right|<\varepsilon\right) \leq P\left(\frac{\left|Y_n-b\right|}{b^2-\varepsilon|b|} \geqslant \varepsilon\right) $$ i cannot figure it out. Thank you very much.

Answer 1

Without loss of generality take $\epsilon <|b|$.

$|b^{2}+b(Y_n-b)|\geq b^{2}-|b(Y_n-b)| \ge b^{2}-\epsilon |b|$ if $|Y_n-b| <\epsilon$. Hence, $\epsilon \le|\frac{Y_n-b}{b^2+b(Y_n-b)}|\le \frac {|Y_n-b|} {b^{2}-\epsilon |b|}$