In a tirangle ABC tan2⁡A2+tan2⁡B2+tan2⁡C2≥K then K==

tan⁡A2−tan⁡B22+tan⁡B2−tan⁡C22+tan⁡C2−tan⁡A22≥0⇒2tan2⁡A2+tan2⁡B2+tan2⁡C2−2tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2≥0⇒tan2⁡A2+tan2⁡B2+tan2⁡C2≥tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2In⁡ triangle ABC,A+B+C=180∘⇒A2+B2+C2=90∘⇒tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1⇒tan2⁡A2+tan2⁡B2+tan2⁡C2≥1

In a tirangle ABC tan2⁡A2+tan2⁡B2+tan2⁡C2≥K then K==

tan⁡A2−tan⁡B22+tan⁡B2−tan⁡C22+tan⁡C2−tan⁡A22≥0⇒2tan2⁡A2+tan2⁡B2+tan2⁡C2−2tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2≥0⇒tan2⁡A2+tan2⁡B2+tan2⁡C2≥tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2In⁡ triangle ABC,A+B+C=180∘⇒A2+B2+C2=90∘⇒tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1⇒tan2⁡A2+tan2⁡B2+tan2⁡C2≥1

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$In a tirangle A B C \tan^{2} \frac{A}{2} + \tan^{2} \frac{B}{2} + \tan^{2} \frac{C}{2} \geq K then K =$ =

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Detailed Solution

$\begin{array}{l} {(\tan \frac{A}{2} - \tan \frac{B}{2})}^{2} + {(\tan \frac{B}{2} - \tan \frac{C}{2})}^{2} + {(\tan \frac{C}{2} - \tan \frac{A}{2})}^{2} \geq 0 \\ \Rightarrow 2 (\tan^{2} \frac{A}{2} + \tan^{2} \frac{B}{2} + \tan^{2} \frac{C}{2}) - 2 (\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2}) \geq 0 \\ \Rightarrow \tan^{2} \frac{A}{2} + \tan^{2} \frac{B}{2} + \tan^{2} \frac{C}{2} \geq \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} \\ In triangle ABC, A + B + C = 180^{\circ} \Rightarrow \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ} \\ \Rightarrow \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1 \\ \Rightarrow \tan^{2} \frac{A}{2} + \tan^{2} \frac{B}{2} + \tan^{2} \frac{C}{2} \geq 1 \end{array}$