Matrix Norm Inequality $\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$ So I'm trying to prove that
$\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$.
I've written the right hand side in terms of rows, but this method doesn't seem to be getting me anywhere. 
Where else should I go?

Writing $A=(A_1,\dots,A_n)^\mathrm{T}$ with $A_i$ being the $i$-th row of the matrix, let $A_j$ be the row for which 
$$
\lVert A\rVert_\infty = \max_{1\leq i\leq n }\lVert A_i\rVert_1 = \lVert A_j\rVert_1 = \sum_{k=1}^n \left|A_{i,j}\right|
$$
Then
$$
n\lVert A\rVert_2^2 = n\sum_{i=1}^n \lVert A_i\rVert_2^2 \geq  n\lVert A_j\rVert_2^2 \geq \lVert A_j\rVert_1^2 = \lVert A\rVert_\infty^2
$$
where the last inequality is "standard" (relation between 1 and 2-norm on $\mathbb{R}^n$, can be proven with Cauchy-Schwarz).
