Proving the inequality $\|A\|_2 ^2 \le \|A\|_1 \|A\|_\infty$

Question

Consider $m\times n$ matrix $A$. I need to show that the following ineqaulity is satisified:

$$\|A\|_2^2 \le \|A\|_1 \|A\|_\infty$$

In all of my attempts I only managed to show that $\|A\|_2 ^2 \le \sqrt n\|A\|_1 \|A\|_\infty$ or similar inequalities, and I could not get rid of the $\sqrt n$. Can someone please give some hints?

Answer 1

$$ \|A\|_2^2=\lambda_{\max}(A^*A)\leq\|A^*A\|_\infty\leq\|A^*\|_\infty\|A\|_\infty=\|A\|_1\|A\|_\infty $$

Answer 2

If $A=(x_{ij})_{1\leqslant i,j\leqslant n}$,$$\|A\|_2^2=\sum_{i,j=1}^n{x_{ij}}^2=\sum_{i,j=1}^n|x_{ij}|^2\leqslant\sum_{i,j=1}^n|x_{ij}|\|A\|_\infty=\|A\|_\infty\|A\|_1.$$