Proof of inequality $P(X \geq \lambda) \leq \frac{\sigma^2}{\sigma^2 + \lambda^2}$ Hello I have the assumptions that $E(X) = 0$ and $\operatorname{Var}(X) = {\sigma}^{2}$ . Why does this inequality hold $P(X \geq \lambda) \leq \frac{{\sigma}^{2}}{{\sigma}^{2} + {\lambda}^{2}}$ for $ \lambda \geq 0$.
It seems like a stronger statement than the Markov inequality and I assume therefore that it must have something to do with the fact that the expectation of $X$ is 0.

This inequality is called Cantelli's inequality (Probability and Measure by Patrick Billingsley, Exercise 5.5).  To prove it, note that for $x > 0$, $\{X \geq \lambda\} \subseteq \{(X + x)^2 \geq (\lambda + x)^2\}$, hence it follows by Markov's inequality that
\begin{align}
P(X \geq \lambda) \leq P((X + x)^2 \geq (\lambda + x)^2) \leq \frac{E[(X + x)^2]}{(\lambda + x)^2} = \frac{\sigma^2 + x^2}{(\lambda + x)^2}. \tag{1}
\end{align}
Observe that (i.e., "completing the squares"):
\begin{align}
\frac{\sigma^2 + x^2}{(\lambda + x)^2} = 
(\lambda^2 + \sigma^2)\left[\frac{1}{\lambda + x} - \frac{\lambda}{\lambda^2 + \sigma^2}\right]^2 + \frac{\sigma^2}{\sigma^2 + \lambda^2},
\end{align}
whence
\begin{align}
\min_{x > 0}\frac{\sigma^2 + x^2}{(\lambda + x)^2} = \frac{\sigma^2}{\sigma^2 + \lambda^2}. \tag{2}
\end{align}
Combining $(1)$ and $(2)$ gives:
\begin{align}
P(X \geq \lambda) \leq \min_{x > 0}\frac{\sigma^2 + x^2}{(\lambda + x)^2}
= \frac{\sigma^2}{\sigma^2 + \lambda^2}.
\end{align}

For a general random variable $X$ with mean $m$ and variance $\sigma^2$, the inequality becomes (simply note $X - m$ has mean $0$ and use the inequality proved for the zero mean case):
\begin{align}
P(X - m \geq \lambda) \leq \frac{\sigma^2}{\sigma^2 + \lambda^2}.
\end{align}
