Proof
First, notice that if \(X > \mu + a\text{,}\) then \(X - \mu \> a\) and so \((x-\mu)^2 > a^2\text{.}\) Similarly for \(X < \mu - a\text{,}\) \((x-\mu)^2 > a^2\text{.}\)
Starting with the definition of variance for a continuous variable X,
\begin{align*}
\sigma^2 & = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) dx\\
& \ge \int_{-\infty}^{\mu-a} (x - \mu)^2 f(x) dx + \int_{\mu + a}^{\infty} (x - \mu)^2 f(x) dx\\
& \ge \int_{-\infty}^{\mu-a} a^2 f(x) dx + \int_{\mu + a}^{\infty} a^2 f(x) dx\\
& = a^2 \left ( \int_{-\infty}^{\mu-a} f(x) dx + \int_{\mu + a}^{\infty} f(x) dx \right )\\
& = a^2 P( X \le \mu - a \text{ or } X \ge \mu + a )\\
& = a^2 P( \big | X - \mu \big | \ge a)
\end{align*}
Dividing by \(a^2\) and taking the complement gives the result.