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,$\left(	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}ight)^2 \leq \frac{1}{27}$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

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

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