Question

In a triangle $ABC$, $\left(\tan \dfrac{A}{2} \tan \dfrac{B}{2} \tan \dfrac{C}{2} \right)^2 \le$

• A. $\dfrac{1}{27}$
• B. $\dfrac{1}{18}$
• C. $\dfrac{1}{9}$
• D. $\dfrac{1}{3}$

Hint:

Use equality case of triangle angle means for maximum/minimum value trigonometric inequalities.

Answer 1

The Correct Option is A

In any triangle $ABC$, maximum value of product $\tan \dfrac{A}{2} \tan \dfrac{B}{2} \tan \dfrac{C}{2}$ is obtained when $A = B = C = \dfrac{\pi}{3}$
Then: $\tan \dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$
So: $\left(\dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{3}} \cdot \dfrac{1}{\sqrt{3}}\right)^2 = \left(\dfrac{1}{3\sqrt{3}}\right)^2 = \dfrac{1}{27}$