Ta có: \(\cot A + \cot B + \cot C = \frac{{\cos A}}{{\sin A}} + \frac{{\cos B}}{{\sin B}} + \frac{{\cos C}}{{\sin C}}\).

Mà áp dụng hệ quả của định lí côsin ta có:

\(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}};\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}};\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\)

\( \Rightarrow \cot A + \cot B + \cot C = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc.\sin A}} + \frac{{{a^2} + {c^2} - {b^2}}}{{2ac.\sin B}} + \frac{{{b^2} + {a^2} - {c^2}}}{{2ab.\sin C}}\left( 1 \right)\)

Ta có: \({S_{ABC}} = \frac{1}{2}bc\sin A = \frac{1}{2}ab\sin C = \frac{1}{2}ab\sin B\left( 2 \right)\)

Kết hợp (1) và (2) ta được:

\(\cot A + \cot B + \cot C = \frac{{{b^2} + {c^2} - {a^2}}}{{4{S_{ABC}}}} + \frac{{{a^2} + {c^2} - {b^2}}}{{4{S_{ABC}}}} + \frac{{{b^2} + {a^2} - {a^2}}}{{4{S_{ABC}}}}\)

\( = \frac{{{b^2} + {c^2} - {a^2} + {c^2} + {a^2} - {b^2} + {a^2} + {b^2} - {c^2}}}{{4{S_{ABC}}}} = \frac{{{a^2} + {b^2} + {c^2}}}{{4{S_{ABC}}}}\)

\( = \frac{{{a^2} + {b^2} + {c^2}}}{{4\frac{{abc}}{{4R}}}} = \frac{{R\left( {{a^2} + {b^2} + {c^2}} \right)}}{{abc}}\) (đpcm)