Show $mathbb{P}[X-malpha]leq frac{sigma^2}{sigma^2+alpha^2}$

Question

I found this problem in an old statistics book:
Suppose $X$ is a square integrable random variable  with mean $m$ and variance $sigma^2$. For any $alpha>0$, show
$$
mathbb{P}[X-m>alpha]leqfrac{sigma^2}{sigma^2 +alpha^2}
$$
At first I thought that the inequality results from  direct application of Markov-Chebyshev's inequality, but when I actually tried it I realized it was not so. Does anybody know about this inequality and how to obtain it?

Oliver Diaz · Answer

That is is known as Cantelli's inequality. It can be obtained from Chebyshev's but with a twist.
For any $x>0$, $alpha+x>0$ and so,
$$ 
begin{align}
mathbf{P}[X-m>alpha]&=mathbf{P}[X-m+x>x+alpha]\
&leq frac{mathbf{E}[(X-m+x)^2]}{(alpha+x)^2}=frac{sigma^2+x^2}{(alpha +x)^2}=:g(x)
end{align}
$$
The game now is to find the best $x$. You can use differential Calculus  to check that $x=frac{sigma^2}{alpha}$ does the trick.

Show $mathbb{P}[X-m>alpha]leq frac{sigma^2}{sigma^2+alpha^2}$

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