In a triangle ABC,left(tan frac{A}{2} tan fra | vedclass

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(A) We know that for $	riangle ABC$,$	an \left(\frac{A}{2}ight) 	an \left(\frac{B}{2}ight) + 	an \left(\frac{B}{2}ight) 	an \left(\frac{C}{

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$\frac{1}{27}$

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(A) We know that for $	riangle ABC$,$	an \left(\frac{A}{2}ight) 	an \left(\frac{B}{2}ight) + 	an \left(\frac{B}{2}ight) 	an \left(\frac{C}{2}ight) + 	an \left(\frac{C}{2}ight) 	an \left(\frac{A}{2}ight) = 1$. Since $A, B, C$ are angles of a triangle,$	an \left(\frac{A}{2}ight), 	an \left(\frac{B}{2}ight), 	an \left(\frac{C}{2}ight) > 0$. Applying the $AM \geq GM$ inequality to these three terms: $\frac{	an \left(\frac{A}{2}ight) 	an \left(\frac{B}{2}ight) + 	an \left(\frac{B}{2}ight) 	an \left(\frac{C}{2}ight) + 	an \left(\frac{C}{2}ight) 	an \left(\frac{A}{2}ight)}{3} \geq \left[	an \left(\frac{A}{2}ight) 	an \left(\frac{B}{2}ight) \cdot 	an \left(\frac{B}{2}ight) 	an \left(\frac{C}{2}ight) \cdot 	an \left(\frac{C}{2}ight) 	an \left(\frac{A}{2}ight)ight]^{\frac{1}{3}}$. Substituting the sum as $1$: $\frac{1}{3} \geq \left[	an^2 \left(\frac{A}{2}ight) 	an^2 \left(\frac{B}{2}ight) 	an^2 \left(\frac{C}{2}ight)ight]^{\frac{1}{3}}$. Cubing both sides: $\frac{1}{27} \geq 	an^2 \left(\frac{A}{2}ight) 	an^2 \left(\frac{B}{2}ight) 	an^2 \left(\frac{C}{2}ight) = \left(	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}ight)^2$. Thus,$\left(	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}ight)^2 \leq \frac{1}{27}$.

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

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