Inequality similar to Paley-Zygmund inequality I cannot do the following exercise:

The term on the right is equal to $$1-\frac{E[X-\lambda]^2}{E[(X-\lambda)^2]}.$$
I'm not sure how to proceed, I thought there was some similarity to the Payley-Zygmund Inequality... however X is not monotone.
 A: This is a general trick to keep in mind when working with concentration inequalities, you introduce a constant and show that some result holds for any value of that constant, and so must hold for the smallest possible value of that constant.
Let $s \in (-\lambda, \infty)$, then note that
\begin{align*}
\mathbb{P}(X \ge \lambda) 
&=\mathbb{P}(X+s \ge \lambda+s)\\ 
&= \mathbb{P}( (X+s)^2 \ge (\lambda+s)^2) \\
&\le \frac{\mathbb{E}(X+s)^2}{(\lambda+s)^2} \\
&= \frac{\sigma^2 + s^2}{(\lambda + s)^2} =: f(s)
\end{align*}
where the inequality holds by Markov's inequality and the fact that $\mathbb{E}(X)=0$. The idea is to make the right hand side as small as possible (the smallest upper bound possible), and note that the right hand side holds for every $s$ in the interval. In other words, we have
$$
\mathbb{P}(X \ge \lambda) \le \inf_{s \in (-\lambda, \infty)} f(s)
$$
Taking the derivative of $f$, it can easily be shown the minimizer of $f(s)$ is
$$
s^* = \frac{\sigma^2}{\lambda}.
$$
Plugging this into the bound yields the desired result. Note that more generally, for any $\lambda \ge 0$
$$
\mathbb{P}(X - \mathbb{E}(X) \ge \lambda ) \le \frac{\text{Var}(X)}{\text{Var}(X) + \lambda^2}.
$$
As an exercise, use the above technique to show:
$$
\mathbb{P}(|X - \mathbb{E}(X)| \ge \lambda ) \le \frac{2\text{Var}(X)}{\text{Var}(X) + \lambda^2}
$$
and compare this to Chebyshev's inequality.
