Suppose $A$ is a $m\times n$ matrix.
Then Prove that, \begin{equation*} \|A\|_2\leq \sqrt{\|A\|_1 \|A\|_{\infty}} \end{equation*}
I have proved the following relations: \begin{align*} \frac{1}{\sqrt{n}}\|A\|_{\infty}\leq \|A\|_2\leq\sqrt{m}\|A\|_{\infty}\\ \frac{1}{\sqrt{m}}\|A\|_{1}\leq \|A\|_2\leq\sqrt{n}\|A\|_{1} \end{align*} Also I feel that somehow Holder's inequality for the special case when $p=1$ and $q=\infty$ might be useful.But I couldn't prove that.
Edit: I would like to have a prove that do not use the information that $\|A\|_2=\sqrt{\rho(A^TA)}$
Usage of inequalities like Cauchy Schwartz or Holder is fine.
$$\|A\|_{2}^2=\sum_{j=1}^{n}\sum_{i=1}^{m}|a|^2_{ij} =\sum_{j=1}^{n}\sum_{i=1}^{m}|a_{ij}||a_{ij}|\leq \sum_{j=1}^{n}\sum_{i=1}^{m}|a_{ij}|\max_{\substack{1\leq i\leq n\\1\leq j\leq n}}{|a_{ij}|} $$ $$=\max_{\substack{1\leq i\leq n\\1\leq j\leq n}}{|a_{ij}|}\sum_{j=1}^{n}\sum_{i=1}^{m}|a_{ij}| =\|A\|_{\infty}\|A\|_{1}.$$
Remark: The same proof works almost identically if the $1-$ norm and $2-$norm are defined by $\|A\|_{1}= \sum_{j=1}^{n}\max_{1\leq i \leq m}{|a_{ij}|}$ and $\|A\|^2_{2}= \sum_{j=1}^{n}\max_{1\leq i \leq m}{|a_{ij}|^2}$ or $\|A\|_{1}= \sum_{i=1}^{m}\max_{1\leq j \leq n}{|a_{ij}|}$ and $\|A\|^2_{2}= \sum_{i=1}^{m}\max_{1\leq j \leq n}{|a_{ij}|^2}$.