Suppose on the contrary that all three numbers $2a - \frac{1}{b}$, $2b - \frac{1}{c}$ and $2c - \frac{1}{a}$ are greater than $2$. Since $abc = 2$, at least one of $a$, $b$, $c$ is positive. WLOG assume $a > 0$. Then $2c - \frac{1}{a} > 2$ implies $c > 1 + \frac{1}{2a} > 0$, and $2b - \frac{1}{c} > 2$ implies $b > 1 + \frac{1}{2c} > 0$.

Next, from $2b - \frac{1}{c} > 2$, we have
$$b > 1 + \frac{1}{2c}. \qquad (1)$$
From $2a - \frac{1}{b} > 2$ and $abc = 2$, we have $\frac{4}{bc} - \frac{1}{b} > 2$, and so
$$b < \frac{2}{c} - \frac{1}{2}. \qquad (2)$$
Combining (1) and (2), we have $1 + \frac{1}{2c} < b < \frac{2}{c} - \frac{1}{2}$, which easily implies $c < 1$. Similarly we can prove that $a < 1$ and $b < 1$. Together with the fact that they are all positive, we have $abc < 1$, contradicting the condition $abc = 2$. Therefore, at most two of $2a - \frac{1}{b}$, $2b - \frac{1}{c}$ and $2c - \frac{1}{a}$ are greater than $2$.