Increasing function, derivative, f' (x) = (MX 2-2x+ 1)/x ≥ 0 holds (x > 0).

That is, when x > 0, the function g (x) = MX 2-2x+ 1 ≥ 0 holds. So m > 0

The symmetry axis of G(x) = 1/m > 0. When x > 0, the minimum value of g(x) = g (1/m) = (m-1)/m ≥ 0, and the solution is m≥ 1.