One sided Chebyshev's inequality

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.

I have a couple of proofs at http://www.se16.info/hgb/cheb.htm#OTProof and http://www.se16.info/hgb/cheb2.htm

One of these, loosely based on Probability and Random Processes by Grimmett and Stirzaker, would give a proof like this:

With $a>0$, for any $b\ge 0$ $$P(X\ge a) = P(X+b \ge a+b) \le E\left[\dfrac{(X+b)^2}{(a+b)^2}\right] = \dfrac{\sigma^2+b^2}{(a+b)^2}$$

But treating $\dfrac{\sigma^2+b^2}{(a+b)^2}$ as a function of $b$, the minimum occurs at $b = \sigma^2 / a$, so $$P(X\ge a) \le \dfrac{\sigma^2+(\sigma^2/a)^2}{(a+\sigma^2/a)^2} =\dfrac{\sigma^2(a^2+\sigma^2)}{(a^2+\sigma^2)^2} = \dfrac{\sigma^2}{ \sigma^2+a^2}.$$