1. Given real numbers $a, b, c$ satisfy $abc=1$, then the number of numbers greater than 1 among $2a-\frac{1}{b}, 2b-\frac{1}{c}, 2c-\frac{1}{a}$ is at most $\qquad$.
1. 2 Detailed Explanation: Assume $2a - \frac{1}{b}$, $2b - \frac{1}{c}$, and $2c - \frac{1}{a}$ are all greater than 1. From $abc = 1$, we know that at least one of $a$, $b$, and $c$ is positive. Without loss of generality, assume $a > 0$, then $2c > 2c - \frac{1}{a} > 1 \Rightarrow c > 0$, thus $2b > 2b - \frac{1}{c} > 1 \Rightarrow b > 0$. Therefore, $a$, $b$, and $c$ are all positive. From $2b - \frac{1}{c} > 1$ $\Rightarrow b > \frac{1}{2}\left(1 + \frac{1}{c}\right)$; $2a - \frac{1}{b} = \frac{2}{bc} - \frac{1}{b} > 1 \Rightarrow b < \frac{2}{c} - 1$; we can get: $\frac{1}{2}\left(1 + \frac{1}{c}\right) < b < \frac{2}{c} - 1 \Rightarrow c < 1$; similarly, $a < 1$, $b < 1$. This contradicts $abc = 1$. Therefore, $2a - \frac{1}{b}$, $2b - \frac{1}{c}$, and $2c - \frac{1}{a}$ cannot all be greater than 1. In fact, $2a - \frac{1}{b}$, $2b - \frac{1}{c}$, and $2c - \frac{1}{a}$ can have 2 of them greater than 1. (For example, take $a = 3$, $b = -2$, $c = -\frac{1}{6}$)