Bài 113123

Question

Cho $a,b,c\in R$.
Chứng minh $\frac{|a-c|}{\sqrt{1+a^2}\sqrt{1+c^2}}\leq \frac{|a-b|}{\sqrt{1+a^2}\sqrt{1+b^2}}+\frac{|b-c|}{\sqrt{1+b^2}\sqrt{1+c^2}}$.

score 1 · Answer 1

Đặt: $a=\tan \alpha, b= \tan \beta, c=\tan \gamma $
Ta có:
$\frac{|a-c|}{\sqrt{a^2+1}. \sqrt{c^2+1} }=\frac{|\tan \alpha - \tan \gamma |}{\frac{1}{|\cos \alpha |}.\frac{1}{|\cos \gamma |} } = |\sin \alpha .\cos \gamma - \cos \alpha . \sin \gamma | $
$=|\sin (\alpha -\gamma )|$
Tương tự: $\frac{|a-b|}{\sqrt{a^2+1}. \sqrt{b^2+1} }=|\sin (\alpha -\beta )|$
$\frac{|b-c|}{\sqrt{b^2+1}. \sqrt{c^2+1} }=|\sin (\beta - \gamma )|$
Ta lại có:
$|\sin (\alpha - \gamma )|= |\sin [(\alpha -\beta )+(\beta -\gamma )]|$
$=|\sin (\alpha -\beta ). \cos (\beta - \gamma )+\cos (\alpha -\beta ). \sin (\beta -\gamma )|$
$ \leq |\sin (\alpha -\beta )|+|\sin (\beta -\gamma )|$
(Do $|\cos (\beta -\gamma )|\leq 1, |\cos (\alpha -\beta )|\leq 1$)
Từ đó ta được:
$\frac{|a-c|}{\sqrt{a^2+1}. \sqrt{c^2+1} }\leq \frac{|a-b|}{\sqrt{a^2+1}. \sqrt{b^2+1} }+\frac{|b-c|}{\sqrt{b^2+1}. \sqrt{c^2+1} }$
Dấu bằng xảy ra $\Leftrightarrow |\cos (\beta -\gamma )|=|\cos (\alpha -\beta )|= 1$
$\Leftrightarrow \begin{cases}\beta -\gamma =k \pi \\ \alpha -\beta =l \pi \end{cases}(k, l \in Z) $
$\Leftrightarrow \tan \alpha = \tan \beta = \tan \gamma \Leftrightarrow a=b=c$

Bài 113123

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Bài 113123

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