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)

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