Show that if $u,v \in \mathbb R^m$, then $|u\cdot v|\leq \|u\|_\infty\|v\|_1$.
By Cauchy-Schwarz, $$|u\cdot v|\leq \|u\|_2\|v\|_2$$ Note also that $\|u\|_1\leq \sqrt{m}\|u\|_2 \leq m\|u\|_\infty$, and $\|v\|_\infty \leq \|v\|_2 \leq \|v\|_1$, so we get
$$|u\cdot v|\leq \sqrt{m}\|u\|_\infty\|v\|_1$$
How to get rid of $\sqrt{m}$ to get the desired inequality?
Let $u=(u_1, \dots, u_m)$ and $v=(v_1, \dots, v_m)$. You have
$$ | u \cdot v | = \left| \sum_{k=1}^m u_k v_k \right| \leq \sum_{k=1}^m |u_k v_k| \leq \|u\|_{\infty} \sum_{k=1}^m |v_k| = \|u\|_{\infty} \|v\|_1 $$
(you can put $\|u\|_{\infty}$ out of the sum because all the $|u_k|$ are less than or equal to $\|u\|_{\infty}$).