How to prove this inequation of matrix norm? Suppose a square matrix $A=(a_{ij})_{n\times n}$ is irreducible. It is given that there exits $i_0$ for $$\sum_{j=1}^{n}{|a_{i_0j}|}<\|A\|_{\infty}$$
Out goal is to prove:  $$\rho(A)< \|A\|_{\infty}$$
I was stuck at the very begining.  $\rho(A)$ is the spectral radius of A. I guess the essence is to prove the equality is not satisfied with the given condition. However I don't know how to prove it. Please help me.
 A: Gershgorin circles:
$$
G_i = \left\{z\in\mathbb{C}:|z-a_{ii}|\leq\sum_{j\neq i}|a_{ij}|\right\}.
$$

Taussky's theorem: Let $A\in\mathbb{C}^{n\times n}$ be irreducible. If $\lambda$ is an eigenvalue of $A$ which lies on the boundary of the union of the Gershgorin discs $G_1,\ldots,G_n$, then $\lambda$ lies on the boundary of each Gershgorin circle $G_1,\ldots,G_n$.

We can apply the Taussky's theorem to prove the following:

Let $A\in\mathbb{C}^{n\times n}$ be irreducible and let
  $$\tag{1}
\sum_{j=1}^n|a_{ij}|\leq\gamma\quad\text{for $i=1,\ldots,n$}
$$
  with a strict inequality for at least one $i_0\in\{1,\ldots,n\}$. Then $\rho(A)<\gamma$.

Proof: Assume that $\rho(A)=|\lambda|=\gamma$ for some eigenvalue $\lambda$ of $A$. By (1), all the Gershgorin circles $G_i$ lie in the closed circle $\Gamma=\{z\in\mathbb{C}:|z|\leq\gamma\}$ and hence $\lambda$ lies on the boundary of $\cup_{i=1}^n G_i$. By Taussky's theorem, $\lambda$ lies on the boundary of each of the Georshorin circle $G_1,\ldots,G_n$. However, this is not possible since (1) holds with the strict inequality for some $i_0$ and thus no boundary point of $G_{i_0}$ is a boundary point of $\Gamma$. Hence, there's no eigenvalue $\lambda$ such that $|\lambda|=\gamma$ and thus $\rho(A)<\gamma$.
Now, set $\gamma=\|A\|_{\infty}$ to get the result.
