@Daminark Btw, I've rewritten your proof, so that it is neat and complete!
We show that $\begin{align}\Vert v\Vert_1\leq\sqrt{n}\Vert v\Vert_2. \end{align}$ We are going to use the Cauchy-Schwartz inequality:
$$
\vert\langle x,y\rangle\vert\leq \Vert x\Vert\cdot \Vert y\Vert,
$$
where $\lVert{.}\rVert$ is a norm induced by an inner product. So we have:
$$
\Vert v\Vert_1=\sum_{i=1}^n\vert v_i\vert=\sum_{i=1}^n1\cdot\vert v_i\vert=\vert\langle 1_v,v^* \rangle \vert,
$$
where $1_v=\{1,\dots,1\}$ and $v^*=\{\vert v_1\vert,\dots,\vert v_n\vert\}$.