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Áp dụng BĐT Cô-si ta có
$\sin^4 x \cos^6 x=108.\frac{\sin^2 x}{2}.\frac{\sin^2 x}{2}.\frac{\cos^2 x}{3}. \frac{\cos^2 x}{3}.\frac{\cos^2 x}{3} \le 108.\left ( \frac{\frac{\sin^2 x}{2}+\frac{\sin^2 x}{2}+\frac{\cos^2 x}{3}+ \frac{\cos^2 x}{3}+\frac{\cos^2 x}{3}}{5} \right )^5=\frac{108}{3125}$
Do đó
$ 0< \int\limits_{0}^{\pi} \sin ^4 x \cos ^6xdx\le\int\limits_{0}^{\pi} \frac{108}{3125}dx< \frac{243\pi}{6250}.$
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