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Trigonometric inequality on an acute triangle

Prove that in any acute triangle ABC the following inequality holds:

    \[\frac{\sin A \sin B}{\cos C} + \frac{\sin B \sin C}{\cos A} + \frac{\sin C \sin A}{\cos B} \geq \frac{9}{2}\]

Solution

Since A+B+C = \pi it holds that

(1)   \begin{equation*} \cos C = -\cos (A +B ) = -\cos A \cos B + \sin A \sin B \end{equation*}

and thus

(2)   \begin{equation*} \frac{\sin A \sin B}{\cos C} = 1 + \frac{\cos A \cos B}{\cos C} = 1 + \frac{\tan C}{\tan A + \tan B} \end{equation*}

Using Nesbitt’s inequality we see that

    \begin{align*} \sum \frac{\sin A \sin B}{\cos C} &= \sum \left ( 1 + \frac{\tan C}{\tan A + \tan B} \right ) \\ &=3 + \sum \frac{\tan C}{\tan A + \tan B} \\ &\geq 3 + \frac{3}{2} \\ &= \frac{9}{2} \end{align*}

Equality holds if-f A=B=C=\frac{\pi}{3}.

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