3. Given real numbers $a, b, c$ satisfy $abc=1$. Then the numbers $2a-\frac{1}{b}, 2b-\frac{1}{c}, 2c-\frac{1}{a}$, at most ( ) of these numbers can be greater than 1.
(A) 3
(B) 2
(C) 1
(D) 0
3. B.

Assume $2a-\frac{1}{b}$, $2b-\frac{1}{c}$, $2c-\frac{1}{a}$ are all greater than 1.
From $abc=1$, we know that at least one of $a$, $b$, $c$ is positive. Without loss of generality, let $a>0$.
Then $2c>2c-\frac{1}{a}>1 \Rightarrow c>0$.
Thus, $2b>2b-\frac{1}{c}>1 \Rightarrow b>0$.
Therefore, $a$, $b$, $c$ are all positive.
From $2b-\frac{1}{c}>1$, we get $b>\frac{1}{2}\left(1+\frac{1}{c}\right)$.
From $2a-\frac{1}{b}>1$, we get $\frac{2}{bc}-\frac{1}{b}>1$, i.e., $b1, 2b-\frac{1}{c}=2>1\right)$.