Minimizing a parameter

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What is the smallest value of positive constant m m that will make m x 1 + ( 1 x ) mx-1+\left( \frac { 1 }{ x } \right) greater than or equal to zero for all positive values of x x ?


The answer is 0.25.

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2 solutions

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

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

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

1 + m x + ( 1 x ) 1 + 2 m x ( 1 x ) = 1 + 2 m -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 m 0 -1+2\sqrt{m}\ge0

Therefore, 2 m 1 2\sqrt{m}\ge1

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

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

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

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

Michael Mendrin
Mar 5, 2014

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

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