Bổ đề: $$\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$$
Chứng minh
$$\begin{array}{l}\cot \left( {A + B} \right) = \dfrac{1}{{\tan \left( {A + B} \right)}} = \dfrac{{1 - \tan A\tan B}}{{\tan A + \tan B}}\\ = \dfrac{{1 - \dfrac{1}{{\cot A\cot B}}}}{{\dfrac{1}{{\cot A}} + \dfrac{1}{{\cot B}}}} = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}\\A + B + C = \pi  \Leftrightarrow A + B = \pi  - C \Leftrightarrow \cot \left( {A + B} \right) = \cot \left( {\pi  - C} \right)\\ \Leftrightarrow \cot \left( {A + B} \right) =  - \cot C\\ \Leftrightarrow \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} =  - \cot C\\ \Leftrightarrow \cot A\cot B - 1 =  - \cot C\left( {\cot A + \cot B} \right)\\ \Leftrightarrow \cot A\cot B + \cot B\cot C + \cot C\cot A = 1\end{array}$$
$$\begin{array}{l}\cot A + \cot B + \cot C = \sqrt 3 \\ \Leftrightarrow {\left( {\cot A + \cot B + \cot C} \right)^2} = 3\\ \Leftrightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C + 2\cot A\cot B + 2\cot B\cot C + 2\cot C\cot A\\ = 3\left( {\cot A\cot B + \cot B\cot C + \cot C\cot A} \right)\\ \Leftrightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C - \cot A\cot B - \cot B\cot C - \cot C\cot A = 0\\ \Leftrightarrow \dfrac{1}{2}\left[ {{{\left( {\cot A - \cot B} \right)}^2} + {{\left( {\cot B - \cot C} \right)}^2} + {{\left( {\cot C - \cot A} \right)}^2}} \right] = 0\\ \Leftrightarrow \cot A = \cot B = \cot C\\ \Leftrightarrow A = B = C\left( {A,B,C < \pi } \right)\end{array}$$
Vậy tam giác $$ABC$$ đều.