高中奥数 2022-03-05

作者: 天目春辉 | 来源:发表于2022-03-05 09:47 被阅读0次

2022-03-05-01

（来源: 数学奥林匹克小丛书第二版高中卷不等式的解题方法与技巧苏勇熊斌证明不等式的基本方法 P027 习题13）

已知 $n$ 个实数 $x_{1},x_{2},\cdots ,x_{n}$ 的算术平均值为 $a$ .证明:
$\sum\limits_{k=1}^{n}\left(x_{k}-a\right)^{2}\leqslant \dfrac{1}{2}\left(\sum\limits_{k=1}^{n}\left|x_{k}-a\right|\right)^{2}.$

证明

先证明 $a=0$ 时的情况.此时 $\sum\limits_{k=1}^{n}x_{k}^{2}= -2\sum\limits_{1\leqslant i<j\leqslant n}x_{i}x_{j}\leqslant 2\sum\limits_{i\neq j}^{n}\left|x_{i}x_{j}\right|$ ,于是 $2\sum\limits_{k=1}^{n}x_{k}^{2}\leqslant \sum\limits_{k=1}^{n}x_{k}^{2}+2\sum\limits_{i\neq j}^{n}\left|x_{i}x_{j}\right|=\left(\sum\limits_{k=1}^{n}\left|x_{k}\right|\right)^{2}$ ,结论成立.

当 $a\ne 0$ 时,令 $y_{k}=x_{k}-a$ ,则 $y_{1},y_{2},\cdots ,y_{n}$ 算术平均值为0,于是 $\sum\limits_{k=1}^{n}x_{k}^{2}\leqslant \dfrac{1}{2}\cdot \left(\sum\limits_{k=1}^{n}\left|y_{k}\right|\right)^{2}$ ,故原不等式成立.

2022-03-05-02

（来源: 数学奥林匹克小丛书第二版高中卷不等式的解题方法与技巧苏勇熊斌证明不等式的基本方法 P027 习题14）

设 $x,y,z\geqslant 0$ ,求证:
$x\left(y+z-x\right)^{2}+y\left(z+x-y\right)^{2}+z\left(x+y-z\right)^{2}\geqslant 3xyz.$
并确定等号成立的条件.

证明

无妨设 $x+y+z=1$ ,则原不等式等价于
$4x^{3}+4y^{3}+4z^{3}-4\left(x^{2}+y^{2}+z^{2}\right)+1\geqslant 3xyz.$
又由于 $x^{3}+y^{3}+z^{3}=1+3xyz-3\left(xy+yz+zx\right)$ , $x^{2}+y^{2}+z^{2}=1-2\left(xy+yz+zx\right)$ ,故原不等式等价于 $xy+yz+zx-\dfrac{9}{4}xyz\leqslant \dfrac{1}{4}$ .

不妨设 $x\geqslant y\geqslant z\geqslant 0$ ,则 $x+y\geqslant \dfrac{2}{3}$ , $z\leqslant \dfrac{1}{3}$ .设 $x+y=\dfrac{2}{3}+\delta$ , $z=\dfrac{1}{3}-\delta$ .其中 $\delta \in \left[0,\dfrac{1}{3}\right]$ .于是
$\begin{aligned} xy+z\left(x+y\right)-\dfrac{9}{4}xyz&=xy+\left(\dfrac{1}{3}-\delta\right)\left(\dfrac{2}{3}+\delta\right)-\dfrac{9}{4}xy\left(\dfrac{1}{3}-\delta\right)\\ &=xy\left(\dfrac{1}{4}+\dfrac{9}{4}\delta\right)+\dfrac{2}{9}-\delta^{2}-\dfrac{1}{3}\delta\\ &\leqslant \left(\dfrac{1}{3}+\dfrac{1}{2}\delta\right)^{2}\cdot \left(\dfrac{1}{4}+\dfrac{9}{4}\delta\right)+\dfrac{2}{9}-\delta^{2}-\dfrac{1}{3}\delta\\ &=\dfrac{3}{16}\delta^{2}\left(3\delta-1\right)+\dfrac{1}{4}\\ &\leqslant 0, \end{aligned}$
因此结论成立.

2022-03-05-03

（来源: 数学奥林匹克小丛书第二版高中卷不等式的解题方法与技巧苏勇熊斌证明不等式的基本方法 P027 习题15）

已知正整数 $n\geqslant 2$ ,实数 $a_{1}\geqslant a_{2}\geqslant \cdots \geqslant a_{n}>0$ , $b_{1}\geqslant b_{2}\geqslant \cdots \geqslant b_{n}>0$ ,并且,有 $a_{1}a_{2}\cdots a_{a}=b_{1}b_{2}\cdots b_{n}$ ; $\sum\limits_{1\leqslant i<j\leqslant n}\left(a_{i}-a_{j}\right)\leqslant \sum\limits_{1\leqslant i<j\leqslant n}\left(b_{i}-b_{j}\right)$ .求证: $\sum\limits_{i=1}^{n}a_{i}\leqslant\left(n-1\right)\sum\limits_{i=1}^{n}b_{i}$ .

证明

当 $n=2$ 时, $\left(a_{1}+a_{2}\right)^{2}-\left(a_{1}-a_{2}\right)^{2}=\left(b_{1}+b_{2}\right)^{2}-\left(b_{1}-b_{2}\right)^{2}$

而 $a_{1}-a_{2}\leqslant b_{1}-b_{2}$ ,故 $a_{1}+a_{2}\leqslant b_{1}+b_{2}$ .

当 $n=3$ 时,
$\begin{aligned} &2b_{1}+2b_{2}+2b_{3}-\left(a_{1}+a_{2}+a_{3}\right)\\ =&b_{1}+2b_{2}+3b_{3}+\left(b_{1}-b_{3}\right)-a_{2}-2a_{3}-\left(a_{1}-a_{3}\right)\\ \geqslant &b_{1}+2b_{2}+3b_{3}-a_{2}-2a_{3}\qquad\qquad\qquad\qquad(*)\\ =&\left(b_{1}-b_{3}\right)+2b_{2}+4b_{3}-\left(a_{2}-a_{3}\right)-3a_{3}\\ \geqslant &2b_{2}+4b_{3}-3a_{3}.\qquad\qquad\qquad\qquad(**) \end{aligned}$
(1)若 $2b_{3}\geqslant a_{3}$ ,则 $2b_{2}+4b_{3}\geqslant 6b_{3}\geqslant 3a_{3}$ ,由 $(**)$ 知结论成立.

(2)若 $b_{2}\geqslant a_{2}$ ,则 $b_{1}+2b_{2}\geqslant 3b_{2}\geqslant 3a_{2}\geqslant a_{2}+2a_{3}$ ,由 $(*)$ 知结论成立.

(3)若 $2b_{3}<a_{3}$ , $b_{2}<a_{2}$ ,则 $b_{1}\geqslant 2a_{1}$ ,故 $2\left(b_{1}+ b_{2}+b_{3}\right)> 2b_{1}>4a_{1}>a_{1}+a_{2}+a_{3}$ ,结论仍成立.

对当 $n\geqslant 3$ 时的一般情况,无妨设 $b_{1}b_{2}\cdots b_{n}=1$ .

如果 $a_{1}\leqslant n-1$ 则 $\sum\limits_{i=1}^{n}a_{i}\leqslant n\left(n-1\right)\leqslant \left(n-1\right)\sum\limits_{i=1}^{n}b_{i}$ ,结论成立.

下设 $a_{1}>n-1$ ,则
$\begin{aligned} \sum\limits_{1 \leqslant i<j \leqslant n}\left(a_{i}-a_{j}\right) &=\sum\limits_{i=1}^{n}(n-2 i+1) a_{i} \\ &=\sum\limits_{i=1}^{n} a_{i}+\sum\limits_{i=1}^{n}(n-2 i) a_{i} \\ & \geqslant \sum\limits_{i=1}^{n} a_{i}+(n-2)\left(a_{1}-a_{n-1}\right)-n a_{n} \\ & \geqslant \sum\limits_{i=1}^{n} a_{i}+\left(a_{1}-a_{n-1}\right)-n a_{n}(n \geqslant 3) \end{aligned}$
$\begin{aligned} \sum\limits_{1 \leqslant i<j \leqslant n}\left(b_{i}-b_{j}\right) &=\sum\limits_{i=1}^{n}(n-2 i+1) b_{i} \\ &=\sum\limits_{i=1}^{n}\left[(n-1) b_{i}+(2-2 i) b_{i}\right] \\ & \leqslant(n-1) \sum\limits_{i=1}^{n} b_{i}-2 b_{2}-2(n-1) b_{n} \end{aligned}$
因此,不妨设 $a_{1}-a_{n-1}-na_{n}+2b_{2}+2\left(n-1\right)b_{n}<0$ ,不然结论显然成立.

于是,有 $na_{n}>2\left(n-1\right)b_{n}+2b_{2}\geqslant 2nb_{n}$ ,故 $a_{n}>2b_{n}$ .又由 $a_{1}a_{2}\cdots a_{n}=1$ 得 $a_{n}\leqslant 1$ ,所以 $a_{1}-\left(n-1\right)a_{n}>\left(n-1\right)-\left(n-1\right)=0$ .故 $2b_{2}<a_{n-1}+a_{n}\leqslant 2a_{n-1}$ ,即 $b_{2}<a_{n-1}$ .

因此, $b_{1}b_{2}\cdots b_{n}=a_{1}a_{2}\cdots a_{n}>2b_{n}\cdot b_{2}\cdot a_{1}a_{2}\cdots a_{n-2}$ ,即有 $b_{1}b_{3}b_{4}\cdots b_{n-1}>2a_{1}a_{2}\cdots a_{n-2}$ .

而 $b_{3}\leqslant b_{2}<a_{n-1}\leqslant a_{n-2}$ , $b_{4}\leqslant b_{3}<a_{n-2}\leqslant a_{n-3}$ , $\cdots$ , $b_{n-1}\leqslant b_{n-2}<a_{3}\leqslant a_{2}$ ,则 $b_{1}>2a_{1}$ .

所以 $(n-1) \sum\limits_{i=1}^{n} b_{i}>2(n-1) a_{1}>n a_{1} \geqslant \sum\limits_{i=1}^{n} a_{i}(n \geqslant 3)$ .结论也成立

2022-03-05-04

（来源: 数学奥林匹克小丛书第二版高中卷不等式的解题方法与技巧苏勇熊斌证明不等式的基本方法 P027 习题16）

设 $x,y,z\in \mathbb{R}^{+}$ ,求证:
$\left(xy+yz+zx\right)\left[\dfrac{1}{\left(x+y\right)^{2}+}+\dfrac{1}{\left(y+z\right)^{2}}+\dfrac{1}{\left(z+x\right)^{2}}\right]\geqslant \dfrac{9}{4}.$

证明

易见,此题等价于证明: $\sum\limits_{c y c} \dfrac{y z}{x(y+z)^{2}} \geqslant \dfrac{9}{4(x+y+z)}$ .

不妨设 $x\geqslant y\geqslant z$ ,且 $x+y+z=1$ .则
$\begin{aligned} \sum\limits_{c y c} \dfrac{4 y z}{x(y+z)^{2}} &=\sum\limits_{c y c} \dfrac{(y+z)^{2}-(y-z)^{2}}{x(y+z)^{2}} \\ &=\sum\limits_{c y c} \dfrac{1}{x}-\sum\limits_{c y c} \dfrac{(y-z)^{2}}{x(y+z)^{2}} \\ &=\sum\limits_{c y c} \dfrac{1}{x} \cdot \sum\limits_{c y c} x-\sum\limits_{c y c} \dfrac{(y-z)^{2}}{x(y+z)^{2}} \\ &=9+\sum\limits_{c y c}\left(\dfrac{\sqrt{z}}{\sqrt{y}}-\sqrt{\dfrac{y}{z}}\right)^{2}-\sum\limits_{c y c} \dfrac{(y-z)^{2}}{x(y+z)^{2}} \\ &=9+\sum\limits_{c y c}\left(\dfrac{(y-z)^{2}}{y z}-\dfrac{(y-z)^{2}}{x(y+z)^{2}}\right) \\ &=9+S, \end{aligned}$
式中, $S=\sum\limits_{cyc}\left(y-z\right)^{2}\cdot \left(\dfrac{1}{yz}-\dfrac{1}{x\left(y+z\right)^{2}}\right))$ .

由于 $x>\dfrac{1}{4}$ ,所以 $\dfrac{1}{yz}\geqslant \dfrac{4}{\left(y+z\right)^{2}}>\dfrac{1}{x\left(y+z\right)^{2}}$ ,

又由于 $\dfrac{1}{xz}-\dfrac{1}{y\left(x+z\right)^{2}}\geqslant 0$ ,

$y\left(x+z\right)^{2}\geqslant xz\left(x+y+z\right)$ ,

$x^{2}\left(y-z\right)+xz\left(y-z\right)+yz^{2}\geqslant 0$ ,

因此 $S\geqslant \left(x-y\right)^{2}\cdot \left(\dfrac{1}{xz}-\dfrac{1}{y\left(x+z\right)^{2}}+\dfrac{1}{xy}-\dfrac{1}{z\left(x+y\right)^{2}}\right)$ ,
$\begin{aligned} &\dfrac{1}{xz}-\dfrac{1}{y\left(x+z\right)^{2}}+\dfrac{1}{xy}-\dfrac{1}{z\left(|x|-y\right)^{2}}\\ =&\dfrac{\left(x+z\right)^{2}-x}{xy\left(x+z\right)^{2}}+\dfrac{\left(x+y\right)^{2}-x}{xz\left(x+y\right)^{2}}\\ \geqslant &\dfrac{\left(x+y\right)^{2}-x+\left(x+z\right)^{2}-x}{xy\left(x+z\right)^{2}}\\ =&\dfrac{2x^{2}+2xy+2xz+y^{2}+z^{2}-2x\left(x+y+z\right)}{xy\left(x+z\right)^{2}}\\ \geqslant& 0, \end{aligned}$
于是结论成立.

高中奥数 2022-03-05

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