[Dr.Lib]Note:Math – Elegant Proof I

16th CanadianMO 1984

任给7个实数,求证必有两个数x、y满足:

\[0 \leq\frac{x – y}{1+x y} \leq\frac{\sqrt{3}}{3} \]

拿到题目后第一反应就是和什么公式好像……结果还真是……

证明:

设七个数为\(a_{1}< a_{2}< a_{3}< a_{4}< a_{5}< a_{6}<a_{7}\),令\(\theta _{i}=arctan(a_{i})\)。

有\( \frac{-\pi}{2}<\theta _{1}<\theta _{2}<\theta _{3}<\theta _{4}<\theta _{5}<\theta _{6}<\theta _{7}<\frac{\pi}{2}\)

则必有\( 0<\theta _{i+1}-\theta _{i}\leq \frac{-\pi}{6} i \in \left [ 1,6 \right ] \)

则\( 0<tan(\theta _{i+1}-\theta _{i})= \frac{tan(\theta _{i+1})-tan(\theta _{i})}{1+tan(\theta _{i})tan(\theta _{i+1})}\leq \frac{\sqrt{3}}{3} \)即\[0 \leq \frac{x – y}{1+x y} \leq \frac{\sqrt{3}}{3} \]

练习:

50th Moscow MO 1987 求证:从任意三个正数或四个实数中总能取两个数x,y满足 \[0 \leq \frac{x – y}{1+x y} \leq 1 \]

3th CMO WC 1988

(1)设正数a,b,c满足 \((a^{2}+b^{2}+c^{2})^{2}>2(a^{4}+b^{4}+c^{4})\)

求证:a,b,c是三角形的三边。

(2)设\((a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+ … +a_{n}^{2})^{2}>(n-1)(a_{1}^{4}+a_{2}^{4}+…+a_{n}^{4}) n\geq3\),求证:\({a_{n}}\)中任意三个数是三角形的三边。

看起来就是各种因式分解。。

证明:

(1)

由题意得\(2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}-a^{4}-b^{4}-c^{4}>0\)

即\((a+b-c)(a+c-b)(b+c-a)(a+b+c)>0\)

不妨设\(a\geq b \geq\ c\)

有\((b+c-a)>0\)即a,b,c是三角形的三边。

(2)

\(n=3\)时由(1)立即可得。

\(n>3\)时

\[(n-1)(a_{1}^{4}+a_{2}^{4}+…+a_{n}^{4}) \]
\[<(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+ … +a_{n}^{2})^{2} \]
\[=(\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+ … +a_{n}^{2})^{2} \]
\[\leq (n-1)(\frac{(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{2}}{4}+\frac{(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{2}}{4}+a_{4}^{4}+…+a_{n}^{4}) (柯西不等式) \]
\[=(n-1)(\frac{(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{2}}{2}+a_{4}^{4}+…+a_{n}^{4})\]

所以

\((a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{2}>2(a_{1}^{4}+a_{2}^{4}+a_{3}^{4})\)

由(1)知\(a_{1},a_{2},a_{3}\)是三角形的三边。

由对称性可知,\({a_{n}}\)中任意三个数是三角形的三边。

【To be continued】

CC BY-SA 4.0 [Dr.Lib]Note:Math – Elegant Proof I by Librazy is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

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