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

Question

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

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2 năm 2021-08-23T18:43:00+00:00 1 Answers 12 views 0

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 $A=B=C$