Minimizing a parameter

This is a clear aplication of the Arithmetic-Geometric Inequality.

1. Let rewrite the sentence as $-1+mx+\left( \frac { 1 }{ x } \right)$ .

2. Applying the Arithmetic-Geometric Inequality, we have:

$-1+mx+\left( \frac { 1 }{ x } \right) \ge -1+2\sqrt { mx\left( \frac { 1 }{ x } \right) } =-1+2\sqrt { m }$

Since we want our sentence greater than or equal to zero, just do:

$-1+2\sqrt{m}\ge0$

Therefore, $2\sqrt{m}\ge1$

Hence we have that $m\ge \frac { 1 }{ 4 } \quad$ or $\quad m\le -\frac { 1 }{ 4 }$ .

But we want the smallest positive integer, then $m\ge \frac { 1 }{ 4 }$

Thus, we conclude that the smallest integer $m$ such that $-1+mx+\left( \frac { 1 }{ x } \right)$ is greater than or equal to zero is $\frac { 1 }{ 4 }=$ $\boxed{0.25}$

$\quad \boxed{Q.E.D.}$

Solve mx - 1 + 1/x = 0 for x. The radical is √(1-4m), which is zero for a double root. Hence, m = 1/4.