Cho $\Delta ABC$ thỏa mãn:  $ab \sin \frac{ C}{ 2} + bc \sin \frac{ A}{ 2} +ac \sin \frac{ B}{ 2} = 2S \sqrt{3}   (1) $
Chứng minh $\Delta ABC$ đều.
Hướng dẫn: Áp dụng công thức tính diện tích tam giác
          $S =  \frac{ 1}{ 2} ab \sin C = \frac{ 1}{ 2} bc \sin A = \frac{ 1}{ 2} ac \sin B$
$(1)     \Leftrightarrow \frac{ 1}{ \cos \frac{ C}{ 2} } +\frac{1 }{\cos \frac{ B}{ 2}  } + \frac{ 1}{ \cos \frac{ A}{ 2} } = 2 \sqrt{3} $
Áp dụng bất đẳng thức Cô si:
           $\cos \frac{ A}{ 2} +\cos \frac{ B}{ 2} +\cos \frac{ C}{ 2}  \geq  3 \sqrt[3]{\cos \frac{ A}{ 2} \cos \frac{ B}{ 2} \cos \frac{ C}{ 2}  } $
          $\frac{ 1}{\cos \frac{ A}{ 2}  } + \frac{ 1}{ \cos \frac{ B}{ 2} } +\frac{1 }{ \cos \frac{C }{ 2} }  \geq 3 \frac{ 1}{\sqrt[3]{\cos \frac{ A}{ 2} \cos \frac{ B}{ 2}\cos \frac{ C}{ 2} }   }  $
nên $\left ( \cos  \frac{ A}{ 2} +\cos \frac{ B}{ 2}+\cos \frac{ C}{ 2}     \right )\left ( \frac{ 1}{ \cos \frac{ A}{2 } } + \frac{1 }{\cos \frac{ B}{ 2}  }+\frac{1 }{ \cos \frac{ C}{ 2} }   \right ) \geq  9   (2)$ dấu đẳng
thức xảy ra khi và chỉ khi $A=B=C$
Chứng minh:  $\cos \frac{A }{ 2} +\cos \frac{ B}{ 2} +\cos \frac{ C}{ 2}  \leq   \frac{ 3 \sqrt{3} }{ 2}   (3)$
Thật vậy, Đặt $T=\cos \frac{A }{ 2} +\cos \frac{ B}{ 2} +\cos \frac{ C}{ 2}$
Ta có: $T=\frac{2}{\sqrt{3} }\left[ {\cos \frac{A}{2}. \frac{\sqrt{3} }{2} +\cos \frac{B}{2}. \frac{\sqrt{3} }{2} } \right]+\sqrt{3}\left[ {\frac{\cos \frac{A}{2} }{\sqrt{3} }.\sin \frac{B}{2}  +\frac{\cos \frac{B}{2} }{\sqrt{3} }.\sin \frac{A}{2}  } \right]   $
$\leq \frac{1}{\sqrt{3} }\left[ {\left( {\cos^2 \frac{A}{2}+\frac{3}{4}} \right) }+ \left( {\cos^2 \frac{B}{2}+\frac{3}{4}} \right)  \right]+ $
$+\frac{\sqrt{3}}{2}\left[ {\left( {\frac{\cos^2 \frac{A}{2} }{3}+\sin^2 \frac{B}{2} } \right) }+ \left ( \frac{\cos ^2 \frac{B}{2} }{3}+ \sin ^2 \frac{A}{2}  \right ) \right]  $
$=\frac{3 \sqrt{3} }{2} $
Dấu bằng xảy ra khi: $A=B=C=60^0$ hay $\Delta ABC$ đều.
Từ $(2)$ và $(3)$ ta được : $\frac{ 1}{\cos \frac{ A}{ 2}  } + \frac{1 }{\cos \frac{ B}{ 2}  } +\frac{ 1}{ \cos \frac{ C}{ 2} } \geq  2 \sqrt{3} $  dấu đẳng thức xảy ra
khi và chỉ khi tam giác $ABC$ đều. Vậy $(1)$ đúng thì tam giác $ABC$ đều.

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