If the expression $[\mathrm{mx}-1+\frac{1}{\mathrm{X}})]$ is non-negative for all positive real x, then the minimum value of m must be

We know that ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\ge 0,\mathrm{\forall }\mathrm{x}\in \mathrm{R},$

if $\mathrm{a}>0\text{ and }{\mathrm{b}}^2-4\mathrm{ac}\le 0$, So

$\mathrm{mx}-1+\frac{1}{\mathrm{x}}\ge 0\text{ or }\frac{{\mathrm{mx}}^2-\mathrm{x}+1}{\mathrm{x}}\ge 0$

or ${\mathrm{mx}}^2-\mathrm{x}+1\ge 0\text{ as }\mathrm{x}>0$

Now, $\begin{array}{l} {\mathrm{mx}}^2-\mathrm{x}+1\ge 0\text{ if }\mathrm{m}>0\text{ and }1-4\mathrm{m}\le 0\\ \Rightarrow \mathrm{m}>0\text{ and }\mathrm{m}\ge 1/4\end{array}$

Thus, the minimum value of m is 1/4.