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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,642

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- #2

- Feb 14, 2012

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$\sin^2 \alpha=\cos^2 \beta+\cos^2 \gamma+\cos^2 \delta \ge 3(\cos \beta \cos \gamma \cos \delta)^{\small \dfrac{2}{3}}$

Multiplying this and three other similar inequalities, we have

$\sin^2 \alpha \sin^2 \beta \sin^2 \gamma \sin^2 \delta=81\cos^2 \alpha \cos^2 \beta \cos^2 \gamma \cos^2 \delta$

Consequently we get $abcd=\sqrt{\tan \alpha \tan \beta \tan \gamma \tan \delta}\ge 3$