Điều kiện cần:
$\Delta ABC$ đều ta có:
$\begin{array}{l}
{m_a} = {m_b} = {m_c} = \frac{{a\sqrt 3 }}{2} = \frac{{2R\sin A\sqrt 3 }}{2} = \sqrt 3 \sin A\\
\Rightarrow \frac{{\sin A}}{{{m_a}}} = \frac{{\sin B}}{{{m_b}}} = \frac{{\sin C}}{{{m_c}}}
= \frac{1}{{\sqrt 3 }}
\end{array}$
$\Rightarrow\frac{\sin
A}{m_a}+\frac{\sin B}{m_b}+\frac{\sin C}{m_c}=\sqrt{3}$
* Điều kiện đủ:
Ta có:
$\begin{array}{l}
m_a^2 = \frac{{{b^2} + {c^2}}}{2} - \frac{{{a^2}}}{4} \Rightarrow 2m_a^2 +
\frac{{{a^2}}}{2} = {b^2} + {c^2}\\
\Rightarrow {a^2} + {b^2} + {c^2} = 2m_a^2 + \frac{{3{a^2}}}{2} \ge 2\sqrt 3 a.{m_a}\\
\Rightarrow \frac{1}{{{m_a}}} \ge \frac{{2\sqrt 3 a}}{{{a^2} + {b^2} + {c^2}}}
\Rightarrow \frac{a}{{{m_a}}} \ge \frac{{2\sqrt 3 {a^2}}}{{{a^2} + {b^2} + {c^2}}}
\end{array}$
Tương tự ta có:
$\begin{array}{l}
\frac{b}{{{m_a}}} \ge \frac{{2\sqrt 3 {b^2}}}{{{a^2} + {b^2} + {c^2}}}\\
\frac{c}{{{m_a}}} \ge \frac{{2\sqrt 3 {c^2}}}{{{a^2} + {b^2} + {c^2}}}
\end{array}$
$ \Rightarrow \frac{a}{{{m_a}}} + \frac{b}{{{m_a}}} + \frac{c}{{{m_a}}} \ge 2\sqrt 3
\Rightarrow \frac{{\sin A}}{{{m_a}}} + \frac{{\sin B}}{{{m_b}}} + \frac{{\sin C}}{{{m_c}}}
\ge \sqrt 3 $
Dấu “=” xảy ra $\left\{ \begin{array}{l}
m_a^2 = \frac{3}{4}{a^2}\\
m_b^2 = \frac{3}{4}{b^2}\\
m_c^2 = \frac{3}{4}{c^2}
\end{array} \right. \Leftrightarrow {a^2} = {b^2} = {c^2}$
$ \Leftrightarrow \Delta ABC$đều.