Can anyone give me a hint how to start this problem? Let $X$ be a random variable such that $P(X \leq 0)=0$ and let $\mu =E(X)$ exist. Show that $P(X \geq 2\mu) ≤ \frac{1}{2}$. I don't really know how to start this one... so any hint will be appreciated.
Hint: For all $t \geq 0$ the inequality $X \geq t\,1_{X \geq t}$ holds almost surely.
$P(X \geq 2\mu) = P\left(z \geq \dfrac{2\mu-\mu}{\sigma}\right)= P\left(z\geq \dfrac{\mu}{\sigma}\right)$, but $P(X \leq 0) = 0 \Rightarrow P\left(z \leq \dfrac{0-\mu}{\sigma}\right) = 0 \Rightarrow P\left(z \geq \dfrac{\mu}{\sigma}\right) = P\left(z \leq -\dfrac{\mu}{\sigma}\right) = 0 \leq \dfrac{1}{2}$.