jmc

Question

Find the minimum value of 9x^2^2 x + 4x x for 0 < x < .

Accepted Answer

The given expression can be rewritten as

9 x sin x + \frac{4}{x sin x}

or

9 y + \frac{4}{y},

where

y = x sin x .

By the AM-GM inequality, we have

\frac{9 y + \frac{4}{y}}{2} \geq 9 y \cdot \frac{4}{y} = 6,

so

9 y + \frac{4}{y} \geq 12,

which is the answer.

To see that the minimum is achievable, recall that equality in AM-GM holds when all the terms are equal. Therefore, we want

9 y = \frac{4}{y},

or

y = x sin x = \frac{2}{3} .

Since

x sin x

is a continuous function of

x,

and

0 sin 0 = 0 < \frac{2}{3}

while

\frac{π}{2} sin \frac{π}{2} = \frac{π}{2} > \frac{2}{3},

the equation

x sin x = \frac{2}{3}

must have a solution in the given interval. Therefore, equality holds for some value of

x .