Lời giải
Đề bài:
Cho $ \begin{cases}\alpha_1,\alpha_2, … , \alpha_n \in (0;\frac{\pi}{2}) , n>3\\\sum\limits_{i=1}^n=\pi \end{cases}$Chứng minh rằng: $(n-\sum\limits_{i=1}^n {\tan^2 \alpha_i} )/(n+ \sum\limits_{i=1}^n {\tan^2 \alpha_i} )\leq \cos \frac{2\pi}{n}$
Lời giải
Xét $ f(x)=\tan^2 x, x \in (0;\frac{\pi}{2})$
$f'(x)=\frac{2\tan x}{\cos^2 x}$
$f”(x)=2.\frac{\frac{1}{\cos^2 x}.\cos^2 x-\tan x(-2\sin x.\cos x)}{\cos^4 x}$
$=2.\frac{1+2\sin^2 x}{\cos^4 x}>0, \forall x \in (0;\frac{\pi}{2})$
Do đó , theo BĐT Jensen
$\frac{1}{n}.\sum\limits_{i = 1}^n {\tan^2 \alpha_i}\geq \tan^2 \left ( \sum\limits_{i = 1}^n {\frac{\alpha_i}{n}} \right )=\tan^2 \frac{\pi}{n}$
$\Leftrightarrow n / \left ( n+\sum\limits_{i = 1}^n {\tan^2 \alpha_i} \right )\leq1/\left ( 1+\tan^2 \frac{\pi}{n} \right )=\cos^2\frac{\pi}{n}$
$\Leftrightarrow 2n / \left ( n+\sum\limits_{i = 1}^n {\tan^2 \alpha_i} \right )-1\leq 2\cos^2\frac{\pi}{n}-1$
$\Leftrightarrow \left ( n-\sum\limits_{i = 1}^n {\tan^2 \alpha_i} \right ) / \left ( n+\sum\limits_{i = 1}^n {\tan^2 \alpha_i} \right )\leq \cos\frac{\pi}{n}$
Dấu $”=”$ xảy ra $\Leftrightarrow \alpha_1=\alpha_2= … =\alpha_n=\frac{\pi}{n} \Rightarrow$ đpcm
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