upper bound of probability with s^2 + a^2 as denominator Let X be a discrete random variable with $E(X) = 0$ and $\sigma^{2}$ = var(X) < $\infty$. Show that $P(X$ $\geq$ $a)$ $\leq$ $\frac{\sigma^{2}}{(\sigma^{2}+a^{2})}$ , for all $a$ $\geq$ $0$.
Please help.

For a constant $b>0,$ define the variable: 
$$Z = (X+b)^2.$$
Then, with $E(X)=0$, we have
$$E(Z) = E(X^2)+b^2=\sigma^2 +b^2.$$
By Markov's inequality
$$P(X\geq a) \leq P[Z\geq (a+b)^2] \leq \frac{E(Z)}{(a+b)^2}=\frac{\sigma^2+b^2}{(a+b)^2}.$$
Choose $b = \sigma^2/a$, which minimizes the bound.
Then 
$$P(X\geq a)  \leq \frac{\sigma^2+\sigma^4/a^2}{(a+\sigma^2/a)^2}= \frac{\sigma^2}{\sigma^2+a^2}.$$
