Lower bound for $\mathbb{P}(|X-\mu| \leq k \sigma)$

Question

Let $X$ be a RV and $\mu,\sigma^2$ be its average and variance, respectively. How can I show

$$\mathbb{P}\left(|X-\mu| \leq k \sigma\right) \geq \frac{k^2 - 1}{k^2}$$

for some $k >1$. Had it been $\mathbb{P}\left(|X-\mu| \geq k \sigma \right)$, maybe I would have used Chebyshev's inequality somehow. Or can I modify the latter to reach this conclusion? I don't see it

Answer 1

First $$ \mathbb{P}\left(|X-\mu| \leq k \sigma\right) = 1- \mathbb{P}\left(|X-\mu| \geq k \sigma\right). $$

On the other hand, by Chebyshev's inequality we have $$ \mathbb{P}\left(|X-\mu| \geq k \sigma\right) \leq \frac{1}{k^2}, $$

putting all togheter:

$$ \mathbb{P}\left(|X-\mu| \leq k \sigma\right) \geq 1- \frac{1}{k^2} = \frac{k^2-1}{k^2}.$$

PD: In Chebyshev's inequality we have a $\leq$ inequality, but in our equiation we have $- \mathbb{P}(\cdots)$ so the inequality change to $\geq$.