1983 AIME Problems/Problem 9

Problem

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$ .

Solution

Let $y=x\sin{x}$ . We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$ .

Since $x>0$ and $\sin{x}>0$ because $0< x<\pi$ , we have $y>0$ . So we can apply AM-GM:

$9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12$

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$ .

Therefore, the minimum value is $\boxed{012}$ (when $x\sin{x}=\frac23$ ; since $x\sin x$ is continuous and increasing on the interval $0 \le x \le \frac{\pi}{2}$ and its range on that interval is from $0 \le x\sin x \le \frac{\pi}{2}$ , by the Intermediate Value Theorem this value is attainable).

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1983 AIME Problems/Problem 9

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