How to prove the one-sided Chebyshev's inequality which states that if $X$ has mean $0$ and variance $\sigma^2$, then for any $a > 0$
$$P(X \geq a) \leq \frac{\sigma^2}{\sigma^2+a^2} \quad?$$
Attempted solution:
I know the Chebyshev's inequality which States that$$P(|X-\mu| \geq a) \leq \frac{\mathrm{Var}(X)}{a^2}~.$$
If I first argue that for any $b > 0$
$$P(X \geq a) \leq P{[(X+b)^2 \geq (a+b)^2]} \\ \begin{align} \implies &P(X\geq a) \leq \frac{E(X+b)^2}{(a+b)^2} \\ &P(X \geq a) \leq \frac{E(X^2)+2E(X)b+b^2}{(a+b)^2} \\ &P(X \geq a) \leq \frac{\sigma^2+ b^2}{(a+b)^2} \end{align}$$
I got the correct answer.