We know that \( \Large ax^{2}+bx+c\ge 0 \) if a > 0 and \( \Large b^{2}-yac \le 0 \).

Now \( \Large mx-1+\frac{1}{x}=0 => \frac{mx^{2}-x+1}{x}\ge 0 \)

=> \( \Large mx^{2}-x+1\ge 0\ and\ x > 0 \)

Now, \( \Large mx^{2}-x+1\ge 0,\)  if m > 0 and \( \Large1-4 m \le 0\)  or if m>o and \( \Large m\ge \frac{1}{4} \)

Thus the minimum value of m is \( \Large \frac{1}{4} \).