Given a matrix $A\in\mathbb{R}^{n\times n}$. How to prove the following statement? $$||A||_{L_1}\leq \sqrt{n}\sigma_{\max}(A),$$ where $||A||_{L_1} = \max_{i\in\{1,...,m\}}||\mbox{col}_i(A)||_1$ and $\sigma_{\max}(A)$ is the largest singular value of $A$.
This conclusion is the Fact 9.8.12 xiv) on Page 573 from the book Matrix Mathematics Theory, Facts, and Formulas (Second Edition) by Dennis S. Bernstein. (I checked the reference therein but didn't get any clue.)
The norm in question is conventionally denoted by $\|\cdot\|_1$. The inequality as it stands does not make sense, because the eigenvalues of $A$ can be non-real and there isn't any "maximum" eigenvalue in general, but I suppose that by $\lambda_\max(A)$ you mean the spectral radius $\rho(A)$ of $A$.
At any rate, the inequality is false. E.g. $$ \left\|\pmatrix{1&1\\ 0&1}\right\|_1=2>\sqrt{2}=\sqrt{2}\rho(A). $$ The inequality can be corrected by requiring that $A$ is normal, or more generally, by changing $\rho(A)$ to $\|A\|_2$, the induced $2$-norm (= the largest singular value) of $A$: \begin{aligned} \|A\|_1 &=\max_{j=1,2,\ldots,n}\|Ae_j\|_1\\ &\le\sup_{\|x\|_2=1}\|Ax\|_1\\ &=\sup_{\|x\|_2=1}\langle|Ax|,e\rangle\\ &\le\sup_{\|x\|_2=1}\|\,|Ax|\,\|_2\|e\|_2\\ &=\sqrt{n}\sup_{\|x\|_2=1}\|Ax\|_2\\ &=\sqrt{n}\|A\|_2. \end{aligned}