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 $A=B=C=\frac{\pi}{3}$ .

Trigonometric inequality on an acute triangle

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