Answer
$m=\frac{1}{4}$
Work Step by Step
Step 1. Let $f(x)=mx-1+\frac{1}{x}$. We need to find the smallest positive value of $m$ which gives $f(x)\geq0$.
Step 2. Letting $f’=0$, we have $m-\frac{1}{x^2}=0$, which gives $x=1/\sqrt m$ (positive only).
Step 3. As $f’’=\frac{2}{x^3}\gt0$, we know the above $x$ will give a local minimum.
Step 4. At $x=1/\sqrt m$, we have $f(1/\sqrt m)=m/\sqrt m -1+\sqrt m=2\sqrt m-1$.
Step 5. We need $f(1/\sqrt m)\geq0$; thus $2\sqrt m-1\geq0$ and $m\geq\frac{1}{4}$, which means that the smallest value is $m=\frac{1}{4}$.
