If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be
If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be
- A
$ - \frac{1}{2}$
- B
$0$
- C
$\frac{1}{4}$
- D
$\frac{1}{2}$
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