Bài 104411

Question

a.Cho $0<x_{1},x_{2},...,x_{n} < \frac{\pi}{2}, \forall n \in N$\$\left\{ \begin{array}{l} \end{array} \right.\left.0,1\right \}$
Chứng minh rằng:
$\frac{\tan x_{1}+\tan x_{2}+...+\tan x_{n} }{n}\geq \tan (\frac{x_{1}+x_{2}+...+x_{n} }{n})$
b.Cho $\triangle ABC$ nhọn.Chứng minh rằng:
*$\tan A+\tan B+\tan C \geq 3 \sqrt {3}$
*$\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\geq \sqrt {3}$

score 0 · Answer 1

a.Xét $f(x)=\tan x,x \in (0,\frac{\pi}{2})$
$f'(x)=\frac{1}{\cos^{2}x}$
$f''(x)=\frac{2\sin x}{\cos^{3}x} >0$
Theo BĐT Jensen ta có:
$\frac{f(x_{1})+f(x_{2})+...+f(x_{n})}{n}\geq f(\frac{x_{1}+x_{2}+...+x_{n} }{n})$
$\Rightarrow \frac{\tan x_{1}+\tan x_{2}+...+\tan x_{n}}{n}\geq \tan(\frac{x_{1}+x_{2}+...+x_{n} }{n})$
Dấu "=" xảy ra $\Leftrightarrow x_{1}=x_{2}=...=x_{n}$.
$\Rightarrow $ (ĐPCM)
$b.\triangle ABC$ nhọn,suy ra:
$A,B,C,\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0,\frac{\pi}{2}) \Rightarrow $ Theo câu a:
$\tan A+\tan B+\tan C \geq 3\tan(\frac{A+B+C}{3})\geq 3 \sqrt {3}$
$\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\geq 3\tan(\frac{A+B+C}{6})\geq\frac{3\sqrt {3}}{3}= \sqrt {3}$
$\Rightarrow $ (ĐPCM)

Bài 104411

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Bài 104411

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