Cho tam giác ABC có số đo ba góc là A, B, C thỏa mãn điều kiện \(\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} = \sqrt 3 \). Tam giác ABC là tam giác gì?

Lời giải

Ta có: \[\frac{A}{2} + \frac{B}{2} = \frac{\pi }{2} - \frac{C}{2}\]

\( \Rightarrow \tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right)\)

\( \Leftrightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} = \cot g\frac{C}{2}\)

\( \Leftrightarrow \left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right)\tan \frac{C}{2} = 1 - \tan \frac{A}{2}\tan \frac{B}{2}\)

\[ \Leftrightarrow \tan \frac{A}{2}\tan \frac{C}{2} + \tan \frac{B}{2}\tan \frac{C}{2} = 1 - \tan \frac{A}{2}\tan \frac{B}{2}\]

 \( \Leftrightarrow \tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2} = 1\)

Áp dụng BĐT (a + b + c)² ³ 3(ab + bc + ca) đối với \(\tan \frac{A}{2},\;\tan \frac{B}{2},\;\tan \frac{C}{2}\), ta được:

\({\left( {\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2}} \right)^2} \ge 3\left( {\tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2}} \right) = 3\)

\[ \Rightarrow \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} \ge \sqrt 3 \]

Mà theo đề ra ta có \(\tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} = \sqrt 3 \) nên dấu bằng xảy ra khi \(\tan \frac{A}{2} = \tan \frac{B}{2} = \tan \frac{C}{2} = \frac{{\sqrt 3 }}{3} \Rightarrow A = B = C = \frac{\pi }{3}\).

Vậy tam giác ABC đều.