Show, that $P(\mu - d\sigma < X < \mu +d\sigma) \ge 1-\frac 1 d^2$ , when $\mu < \infty$ and $\sigma^2>0$ and $ d>1$
My Idea:
$P(\mu - d\sigma < X < \mu +d\sigma)$ = $1-P(X \ge \mu+d\sigma) + P(X>\mu-d\sigma) \ge 1- \frac {E[X^2]}{\mu+d\sigma} + P(X>\mu-d\sigma) \ge 1- \frac {E[X^2]}{\mu+d\sigma} = 1- \frac {\mu^2+\sigma^2}{(\mu+d\sigma)^2}$
I used the Chebyshev inequality and the formula for variance, but don't know how to proceed from here. Did I make a mistake? Or is it so obvious, that I just don't recognize it?
$$P(|X-\mu| \le d\sigma) = 1 - P(|X-\mu| > d\sigma)$$
$$P(|X-\mu| > d\sigma) \le \frac{\sigma^2}{d^2 \sigma^2} \quad \text{Chebyshev Inequality}$$