Let X be a random variable such that $X(\omega) \in [0,1] \;\forall \omega\in\Omega$. Show that $$\text{if }\; \mathbb{E}(X)\geq \frac{1}{2}\text{, then }\, \mathbb{P}(X\leq \frac{1}{4})\leq \frac{2}{3}$$
The problem looks like an application of Markov's Inequality, but I couldn't get $\frac{2}{3}$ in the bound. I guess I'm missing something.
$$ \mathsf{P}(1-X\ge 3/4)\le \frac{4}{3}(1-\mathsf{E}X)\le \frac{4}{3}\times \frac{1}{2}=\frac{2}{3}. $$