Let $B_n$ be a brownian motion, apply Doob's inequality to show that:
$$\mathbb E\left(\sup_{\sigma \leq t \leq \tau} \left(\frac{B_t}{t}\right)^2\right)\leq \frac{4\tau}{\sigma^2}.$$
I know a result about Doob's inequality:
$$\mathbf{P} \left[ \sup_{0 \leq t \leq T} B_{t} \geq C \right] \leq \exp \left( - \frac{C^2}{2T} \right).$$
But I don't know how to apply.