Bài 106747

Question

Cho $A, B, C$ là ba góc của một tam giác. Hãy chứng minh rằng :
$\tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan\frac{A}{2} = 1\,\,\,\,\,(1)$
Và $\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2} \le \frac{1}{{3\sqrt 3 }}\,\,\,\,\,(2)$ Dấu bằng trong ($2$) xảy ra khi nào ?

score 0 · Answer 1

$\begin{array}{l}
\,\,\frac{A}{2} + \frac{B}{2} = \frac{\pi }{2} - \frac{C}{2} \Leftrightarrow \tan \left( {\frac{A}{2}

+ \frac{B}{2}} \right) = \cot \frac{C}{2}\\
\Rightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} =

\frac{1}{{\tan \frac{C}{2}}}\\
\Rightarrow \tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan

\frac{C}{2}\tan \frac{A}{2} = 1\,\,\,
\end{array}$
Theo bất đẳng thức Cauchy :
$\begin{array}{l}
1 = \tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan
\frac{A}{2} \ge 3\sqrt[3]{{{{\left( {\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}}
\right)}^2}}}\\
\Rightarrow 1 \ge 27{\left( {\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}} \right)^2}
\Rightarrow \tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2} \le \frac{1}{{3\sqrt 3 }}
\end{array}$
Đẳng thức xảy ra khi:
$\tan \frac{A}{2}\tan \frac{B}{2} = \tan \frac{B}{2}\tan \frac{C}{2} = \tan \frac{A}{2}\tan
\frac{C}{2}$
$ \Rightarrow \tan \frac{A}{2} = \tan \frac{B}{2} = \tan \frac{C}{2} \Leftrightarrow A = B = C$
Suy ra tam giác $ABC$ đều.

Bài 106747

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Bài 106747

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