cho tam giác ABC ,tìm Min P=tan(A\2) + tan(B\2) + tan(C\2)

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1. Áp dụng bất đẳng thức $(a+b+c)^2\ge 3(ab+bc+ca)$

   Ta đi chứng minh $\tan \dfrac{A}{2}\tan \dfrac{B}{2} + \tan \dfrac{B}{2}\tan \dfrac{C}{2} + \tan \dfrac{C}{2}\tan \dfrac{A}{2} = 1$

   $\begin{array}{l} \dfrac{{A + B + C}}{2} = \dfrac{\pi }{2}\\  \Rightarrow \left( {\dfrac{{A + B}}{2}} \right) = \left( {\dfrac{\pi }{2} – C} \right)\\  \Rightarrow \tan \left( {\dfrac{A}{2} + \dfrac{B}{2}} \right) = \tan \left( {\dfrac{\pi }{2} – C} \right) = \dfrac{1}{{\tan C}}\\  \Rightarrow \dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{B}{2}}}{{1 – \tan \dfrac{A}{2}\tan \dfrac{B}{2}}} = \dfrac{1}{{\tan \dfrac{C}{2}}}\\  \Rightarrow \tan \dfrac{A}{2}\tan \dfrac{B}{2} + \tan \dfrac{B}{2}\tan \dfrac{C}{2} + \tan \dfrac{C}{2}\tan \dfrac{A}{2} = 1\\  \Rightarrow {\left( {\tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan \dfrac{C}{2}} \right)^2} \ge 3.1 \Rightarrow \tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan \dfrac{C}{2} \ge \sqrt 3 \\ \end{array}$

   Dấu bằng xảy ra khi và chỉ khi $A=B=C$

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