Cho tam giác ABC với \(BC = a,AC = b,AB = c\), p là nửa chu vi tam giác ABC.
Chứng minh rằng \(abc\left( {\cos A + \cos B + \cos C} \right) = {a^2}\left( {p - a} \right) + {b^2}\left( {p - b} \right) + {c^2}\left( {p - c} \right)\).
Định lý cosin: Cho tam giác ABC có \(AB = c,BC = a,AC = b\) thì \({a^2} = {b^2} + {c^2} - 2bc\cos A\).
Ta có: \(abc\left( {\cos A + \cos B + \cos C} \right) = abc\left( {\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right)\)
\( = a.\frac{{{b^2} + {c^2} - {a^2}}}{2} + b\frac{{{a^2} + {c^2} - {b^2}}}{2} + c\frac{{{a^2} + {b^2} - {c^2}}}{2}\)
\( = \frac{1}{2}\left( {a{b^2} + a{c^2} - {a^3} + {a^2}b + b{c^2} - {b^3} + {a^2}c + {b^2}c - {c^3}} \right)\)
\( = \frac{{{a^2}}}{2}\left( {b + c - a} \right) + \frac{{{b^2}}}{2}\left( {c + a - b} \right) + \frac{{{c^2}}}{2}\left( {a + b - c} \right)\)
\( = {a^2}\left( {p - a} \right) + {b^2}\left( {p - b} \right) + {c^2}\left( {p - c} \right)\)a
