Is it possible to show $P(|X-μ| \geq \alpha)\leq\frac{2\sigma^2}{\sigma^2 + \alpha^2}$ from Cantelli's inequality?

I know that Cantelli's Inequality states that for a random variable $X$ with mean $μ$ and variance $\sigma^2$:

$$P(X-μ\geq \alpha)\leq\frac{\sigma^2}{\sigma^2 + \alpha^2}$$

Here, I am trying to figure out how to show that:

$$P(|X-μ| \geq \alpha)\leq\frac{2\sigma^2}{\sigma^2 + \alpha^2}$$

Is there an easy way to do this directly from Cantelli's inequality or would it require a separate proof? thanks.

Apply your known formula (Cantelli's inequality) twice, once on $X$ and once on $Y=-X$ to get the result. The random variable $Y=-X$ has mean $μ_Y=Ε[Y]=-Ε[Χ]=-μ$ and variance $σ_Y=Var(Y)=(-1)^2Var(X)=σ^2$. So \begin{align}P(|X-μ|\ge α)&=P(X-μ\ge α)+P(X-μ\le -α)\\[0.2cm]&=P(X-μ \ge α)+P(-X-(-μ)\ge α)\\[0.2cm]&=P(X-μ\ge α)+P(Y-μ_Y\ge α)\end{align}
• Hi, thanks for the response. I am just beginning to take probability so sorry if this sounds stupid but what allows us to get the first step above, $P(|X-μ|\ge α)=P(X-μ\ge α)+P(Χ-μ\le -α)$? The farthest I got was that $P(|X-μ|\ge α) = P(X-μ\ge α \lor Χ-μ\le -α)$ but dont know why you can add the inside, since I thought you could only add events when they are disjoint. How do I know these events are disjoint so to say? Thank you so much! – user136503 Jan 29 '16 at 15:03
• $P(A\cup B)=P(A)+P(B)$ when the events $A$ and $B$ are disjoint. You certainly have had this formula before going to random variables? So, here the events $X-μ\ge α$ and $Χ-μ\le -α$ are disjoint, hence the probability of their union is the sum of their probabilities. – Jimmy R. Jan 29 '16 at 15:08
• Yes, there are ways with $X(ω)$ and so on, but drawing number lines is also the way I think about it. – Jimmy R. Jan 29 '16 at 15:16