Ta đi chứng minh $\tan \dfrac{A}{2}.\tan \dfrac{B}{2} + \tan \dfrac{B}{2}.\tan \dfrac{C}{2} + \tan \dfrac{C}{2}.\tan \dfrac{A}{2} = 1$
Ta có: $\begin{gathered} \dfrac{{A + B}}{2} = \dfrac{{\pi - C}}{2} \Rightarrow \tan \left( {\dfrac{{A + B}}{2}} \right) = \tan \left( {\dfrac{{\pi - C}}{2}} \right) \hfill \\ \Rightarrow \dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{B}{2}}}{{1 - \tan \dfrac{A}{2}.\tan \dfrac{B}{2}}} = \dfrac{1}{{\tan \dfrac{C}{2}}} \hfill \\ \Rightarrow \left( {\tan \dfrac{A}{2} + \tan \dfrac{B}{2}} \right).\tan \dfrac{C}{2} = 1 - \tan \dfrac{A}{2}.\tan \dfrac{B}{2} \hfill \\ \Rightarrow \tan \dfrac{A}{2}\tan \dfrac{B}{2} + \tan \dfrac{B}{2}\tan \dfrac{C}{2} + \tan \dfrac{C}{2}.\tan \dfrac{A}{2} = 1 \hfill \\ \end{gathered}$
Áp dụng bất đẳng thức $(a+b+c)^2\ge 3(ab+bc+ca)$ đối với $\tan\dfrac{A}{2},\tan\dfrac{B}{2},\tan\dfrac{C}{2}$ được
$\begin{array}{l}
{\left( {\tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan \dfrac{C}{2}} \right)^2} \ge 3\left( {\tan \dfrac{A}{2}\tan \dfrac{B}{2} + \tan \dfrac{B}{2}\tan \dfrac{C}{2} + \tan \dfrac{C}{2}\tan \dfrac{A}{2}} \right) = 3\\
\Rightarrow \tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan \dfrac{C}{2} \ge \sqrt 3
\end{array}$
Vậy mà theo đề ta có $\tan\dfrac{A}{2}+\tan\dfrac{B}{2}+\tan\dfrac{C}{2}=\sqrt{3}$ nên dấu bằng xảy ra khi $\tan\dfrac{A}{2}=\tan\dfrac{B}{2}=\tan\dfrac{C}{2}\Rightarrow A=B=C=\dfrac{\pi}{3}$
Vậy tam giác ABC đều.
