From R2016: Probabilistic Markov's inequality, we have the inequality \begin{equation} \frac{1}{\lambda} \mathbb{E}(X) \geq \mathbb{P}(X \geq \lambda) \end{equation} Multiplying both sides by $\lambda$, we then conclude \begin{equation} \mathbb{E}(X) \geq \lambda \mathbb{P}(X \geq \lambda) \end{equation} $\square$