Proving $\frac{1}{n}\kappa_2(A)\leq\kappa_1(A)\leq n\kappa_2(A)$ I am trying to show the following inequality is true: $\frac{1}{n}\kappa_2(A)\leq\kappa_1(A)\leq n\kappa_2(A)$. 
Here we denote $\kappa(A)$ to be the condition number of the matrix $A\in\mathbb{R}^{n\times n}$, where $$\kappa(a)=\|A\|\|A^{-1}\|.$$
Beginning with the left-hand-side, $$\frac{1}{n}\kappa_2(A)=\frac{1}{n}\left|\frac{\lambda_1}{\lambda_n}\right|, \ \ \text{where} \ |\lambda_1|\geq...\geq|\lambda_n|>0 \ \text{(theorem)}.$$ I do not see how the first inequality, $\frac{1}{n}\kappa_2(A)\leq\kappa_1(A)$, is achieved. Any advice is appreciated.

I assume that by $\kappa_p(A)$ you mean the condition number coming from the usual matrix norm
$$ ||A||_p=\sup_{x\neq 0}\frac{||Ax||_p}{||x||_p}$$
on $\mathbb{R}^{n\times n}$. One can show that 
$$\frac{1}{\sqrt{n}}||B||_2\leq ||B||_1\leq \sqrt{n}||B||_2$$
for any $B\in \mathbb{R}^{n\times n}$, and using these inequalities with $B=A$ and $B=A^{-1}$ yields the desired result.
