According to result
R4145: Binary union is an upper bound to both sets in the union, we have the inclusion
\begin{equation}
\{ X > \lambda \}
\subseteq \{ X > \lambda \} \cup \{ X = \lambda \}
= \{ X \geq \lambda \}
\end{equation}
Now, applying results
we conclude with
\begin{equation}
\mathbb{P}(X > \lambda)
\leq \mathbb{P}(X \geq \lambda)
\leq \frac{1}{\lambda} \mathbb{E}(X)
\end{equation}
$\square$