If the expression [mx−1+1X)] is non-negative for all positive real x, then the minimum value of m must be

We know that ax2+bx+c≥0,∀x∈R,if a>0 and b2−4ac≤0, Somx−1+1x≥0 or mx2−x+1x≥0or mx2−x+1≥0 as x>0Now,          mx2−x+1≥0 if m>0 and 1−4m≤0⇒ m>0 and m≥1/4Thus, the minimum value of m is 1/4.