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-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

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

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