In Δ ABC, the minimum value of ∑cot2A2.cot2B2∏cot2A2 is

Σcot2A2.cot2B2πcot2A2=tan2A2+tan2B2+tan2C2Now we have (tanA2−tanB2)2≥0tan2A2+tan2B2≥2tan2A2tanB2Similarly write this inequality for other angles and add all 3 in equalities, we get Σ tan2A2≥Σ tanA2tanB2=1

In Δ ABC, the minimum value of ∑cot2A2.cot2B2∏cot2A2 is

Σcot2A2.cot2B2πcot2A2=tan2A2+tan2B2+tan2C2Now we have (tanA2−tanB2)2≥0tan2A2+tan2B2≥2tan2A2tanB2Similarly write this inequality for other angles and add all 3 in equalities, we get Σ tan2A2≥Σ tanA2tanB2=1

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In $Δ A B C$ , the minimum value of $\frac{\sum \cot^{2} \frac{A}{2} . \cot^{2} \frac{B}{2}}{\prod \cot^{2} \frac{A}{2}}$ is

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Detailed Solution

$\frac{Σ \cot^{2} \frac{A}{2} . \cot^{2} \frac{B}{2}}{π \cot^{2} \frac{A}{2}} = \tan^{2} \frac{A}{2} + \tan^{2} \frac{B}{2} + \tan^{2} \frac{C}{2}$

Now we have ${(\tan \frac{A}{2} - \tan \frac{B}{2})}^{2} \geq 0$

$\tan^{2} \frac{A}{2} + \tan^{2} \frac{B}{2} \geq 2 \tan^{2} \frac{A}{2} \tan \frac{B}{2}$

Similarly write this inequality for other angles and add all 3 in equalities, we get $Σ \tan^{2} \frac{A}{2} \geq Σ \tan \frac{A}{2} \tan \frac{B}{2} = 1$