Chapter 8, Problem 8.2TE

If X has mean $\mu$ and standard deviation $\sigma$, the ratio $r\equiv \frac{\left|\mu \right|}{\sigma }$ is called the measurement signal-to-noise ratio of X. The idea is that X can be expressed as $X=\mu +\left(X-\mu \right)$, with $\mu$ representing the signal and $X-\mu$ the noise. If we define $\left|\frac{\left(X-\mu \right)}{\mu }\right|\equiv D$ as the relative deviation of X from its signal (or mean) $\mu$, show that for $a>0$, $P\left\{D\le a\right\}\ge 1-\frac{1}{r^2{a}^2}$