Spectral Radius of a non negative matrix

Question

Let $A \in \mathbb{R}^{n\times n}$ be a non-negative square matrix and $x \in \mathbb{R}^{n}$ be a positive eigenvector. Prove that \begin{gather*} \rho(A)=\max\limits_{x>0}\; \min\limits_{i=1,2,\dots,n}\frac{1}{x_i}\sum_{j=1}^{n}a_{ij}x_j=\min\limits_{x>0}\; \max\limits_{i=1,2,\dots,n}\frac{1}{x_i}\sum_{j=1}^{n}a_{ij}x_j. \end{gather*} using the inequality \begin{gather*} \min\limits_{i=1,2,\dots,n}\left\{\frac{1}{x_i}\sum_{j=1}^{n}a_{ij}x_j\right\}\le \rho(A) \le \max\limits_{i=1,2,\dots,n}\left\{\frac{1}{x_i}\sum_{j=1}^{n}a_{ij}x_j\right\} \end{gather*} I know that I need to use the fact that $x$ is a positive eigenvector, hence its corresponding eigenvalue is $\rho(A)$, I just can't grasp why the max and the min are used on the equality that needs to be proven.