Matrix Norm Bounds

Question

A natural consideration of matrix norms is to compare them, and one of the many standard results on the induced 1, 2, and $\infty$-norms indicate that $$\frac{1}{\sqrt{n}}\|A\|_\infty\leq \|A\|_2\leq \sqrt{m}\|A\|_\infty$$ $$\frac{1}{\sqrt{m}}\|A\|_1\leq \|A\|_2\leq \sqrt{n}\|A\|_1$$ This is proven using Cauchy-Schwarz (among other things). I was recently told that for a square matrix, $$\|A\|_2^2\leq \|A\|_1\|A\|_\infty,$$ which seems more intuitively interesting to me as it's not dependent on the size of your matrix. I tried proving this similarly to the above two by seeing how $\|Ae\|_2$ can get bounded, but wasn't making any headway because when splitting off $\|e\|_2$, it would then be dependent on the matrix size (and not what I'm trying to do). Using simply the definitions, I'm not sure how I could relate the max singular value to the max row and column sums.

Answer 1

Hint: $\|A\|_2^2=\rho(A^\ast A)$ and the spectral radius is a lower bound to all (submultiplicative) matrix norms.