Ta có:
\(\begin{array}{l}m_a^2 = \dfrac{{{b^2} + {c^2}}}{2} - \dfrac{{{a^2}}}{4}\\m_b^2 = \dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4}\\m_c^2 = \dfrac{{{b^2} + {a^2}}}{2} - \dfrac{{{b^2}}}{4}\\ \Rightarrow m_a^2 + m_b^2 + m_c^2 = \dfrac{{{b^2} + {c^2}}}{2} - \dfrac{{{a^2}}}{4} + \dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4} + \dfrac{{{b^2} + {a^2}}}{2} - \dfrac{{{b^2}}}{4}\\ = \dfrac{{{b^2} + {c^2} + {c^2} + {a^2} + {a^2} + {b^2}}}{2} - \dfrac{{{a^2} + {b^2} + {c^2}}}{4}\\ = {a^2} + {b^2} + {c^2} - \dfrac{{{a^2} + {b^2} + {c^2}}}{4}\\ = \dfrac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right)\,\left( 1 \right)\end{array}\)
Mà
\(\begin{array}{l}S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \\ \le \sqrt {p.{{\left( {\dfrac{{p - a + p - b + p - c}}{3}} \right)}^3}} \\ = \sqrt {\dfrac{{a + b + c}}{2}.{{\left( {\dfrac{{\dfrac{{3\left( {a + b + c} \right)}}{2} - \left( {a + b + c} \right)}}{3}} \right)}^3}} \\ = \sqrt {\dfrac{{a + b + c}}{2}.\dfrac{{{{\left( {a + b + c} \right)}^3}}}{{{6^3}}}} \\ = \dfrac{{\sqrt 3 {{\left( {a + b + c} \right)}^2}}}{{36}}\end{array}\)
Áp dụng BĐT Bunhia ta có:
\(\begin{array}{l}\dfrac{{\sqrt 3 {{\left( {a + b + c} \right)}^2}}}{{36}}\\ \le \dfrac{{\sqrt 3 }}{{36}}.\left( {{a^2} + {b^2} + {c^2}} \right)\left( {{1^2} + {1^2} + {1^2}} \right)\\ = \dfrac{{\sqrt 3 }}{{36}}.\left( {{a^2} + {b^2} + {c^2}} \right).3\\ = \dfrac{{\sqrt 3 }}{{12}}.\left( {{a^2} + {b^2} + {c^2}} \right)\end{array}\)
Suy ra \(S \le \dfrac{{\sqrt 3 }}{{12}}\left( {{a^2} + {b^2} + {c^2}} \right) \Leftrightarrow {a^2} + {b^2} + {c^2} \ge \dfrac{{12S}}{{\sqrt 3 }}\)
Thay vào (1) được \(m_a^2 + m_b^2 + m_c^2 = \dfrac{3}{4}\left( {{a^2} + {b^2} + {c^2}} \right) \ge \dfrac{3}{4}.\dfrac{{12S}}{{\sqrt 3 }} = 3\sqrt 3 S\) (đpcm)
Dấu “=” xảy ra khi \(a = b = c\) hay tam giác đều.
