If $X$ is a nonnegative random variable and $a \gt 0$ then, $Pr[X \ge a]$ $\le$ $\frac{E(X)}{a}$.
Proof is like follow. First of all, $E(X | X \ge a)$ $\ge$ $a$ because condition means that varaible is greater or equal than $a$. Simillary, $E(X | X \lt a)$ $\ge$ $0$ because $X$ is a nonnegative random variable. Then, $E(X)$ $=$ $Pr[X \lt a] \cdot E(X | X \lt a)$ $+$ $Pr[X \ge a] \cdot E(X | X \ge a)$ $\ge$ $Pr[X \ge a] \cdot E(X | X \ge a)$ $\ge$ $Pr[X \ge a] \cdot a$. Therefore, claim holds