奥数自学研究

高中奥数 2022-02-14

2022-02-14  本文已影响0人  天目春辉

2022-02-14-01

(来源: 数学奥林匹克小丛书 第二版 高中卷 数列与数学归纳法 冯志刚 习题二 P095 习题07)

(Jenson不等式)设f\left(x\right)\left[a,b\right]上的凸函数(即对任意xy\in \left[a,b\right],都f\left(\dfrac{x+y}{2}\right)\leqslant \dfrac{1}{2}\left(f\left(x\right)+f\left(y\right)\right)).

证明:对任意n个数x_{1},\cdots,x_{n}\in \left[a,b\right],都有
f\left(\dfrac{x_{1}+\cdots+x_{n}}{n}\right)\leqslant \dfrac{1}{n}\left(f\left(x_{1}\right)+\cdots+f\left(x_{n}\right)\right).

证明

对比第10节中平均值不等式的证明二,用其中出现的方法来证这个应用广泛的Jenson 不等式.

n=1,2时,不等式显然成立.

现设不等式对n=2^{k}\left(k\in \mathbb{N}^{*}\right)成立,则由f的定义,可知
\begin{aligned} f\left(\dfrac{x_{1}+\cdots+x_{2^{k+1}}}{2^{k+1}}\right) & \leqslant \dfrac{1}{2}\left(f\left(\dfrac{x_{1}+\cdots+x_{2^{k}}}{2^{k}}\right)+f\left(\dfrac{x_{2^{k}+1}+\cdots+x_{2^{k+1}}}{2^{k}}\right)\right) \\ & \leqslant \dfrac{1}{2}\left(\dfrac{1}{2^{k}} \sum\limits_{j=1}^{2^{k}} f\left(x_{j}\right)+\dfrac{1}{2^{k}} \sum\limits_{j=1}^{2^{k}} f\left(x_{2^{k}+j}\right)\right) \\ &=\dfrac{1}{2^{k+1}} \sum\limits_{j=1}^{2^{k+1}} f\left(x_{j}\right) \end{aligned}
因此,不等式对任意n=2^{k}\left(k\in \mathbb{N}^{*}\right)都成立.

对一般的n\in \mathbb{N}^{*}\left(n\geqslant 3\right),设2^{k}\leqslant n<2^{k+1},k\in \mathbb{N}^{*},记A=\dfrac{1}{n}\left(x_{1}+\cdots+x_{n}\right),则由不等式对2^{k+1}成立,知
f\left(\dfrac{x_{1}+\cdots+x_{n}+\left(2^{k+1}-n\right)A}{2^{k+1}}\right)\leqslant \dfrac{1}{2^{k+1}}\left(\sum\limits_{j=1}^{n}f\left(xj\right)+\left(2^{k+1}-n\right)f\left(A\right)\right),
\dfrac{1}{2^{k+1}}\left(x_{1}+\cdots+x_{n}+\left(2^{k+1}-n\right)A\right)=\dfrac{1}{2^{k+1}}\left(nA+\left(2^{k+1}-n\right)A\right)=A,于是,我们有
2^{k+1}f\left(A\right)\leqslant \sum\limits_{j=1}^{n}f\left(x_{j}\right)+\left(2^{k+1}-n\right)f\left(A\right),
f\left(A\right)\leqslant \dfrac{1}{n}\sum\limits_{j=1}^{n}f\left(x_{j}\right),即不等式对n成立.

命题获证.

2022-02-14-02

(来源: 数学奥林匹克小丛书 第二版 高中卷 数列与数学归纳法 冯志刚 习题二 P095 习题08)

设正实数x_{1},\cdots ,x_{n}满足x_{1}+\cdots+x_{n}=1,这里n\in \mathbb{N}^{*},n\geqslant 2.

证明:\prod\limits_{k=1}^{n}\left(1+\dfrac{1}{x_{k}}\right)\geqslant\prod\limits_{k=1}^{n}\left(\dfrac{n-x_{k}}{1-x_{k}}\right).

证明

引理f\left(x\right)\left(0,1\right)上的凸函数,n\in \mathbb{N}^{*},n\geqslant 2,正实数x_{1},\cdots ,x_{n}满足x_{1}+\cdots+x_{n}=1,则
\sum\limits_{i=1}^{n}f\left(x_{i}\right)\geqslant \sum\limits_{i=1}^{n}f\left(\dfrac{1-x_{i}}{n-1}\right).
引理的证明:由Jenson不等式,知
\begin{aligned} \sum\limits_{i=1}^{n}f\left(x_{i}\right)&=\sum\limits_{i=1}^{n}\left(\dfrac{1}{n-1}\sum\limits_{j\neq i}f\left(x_{j}\right)\right)\\ &\geqslant \sum\limits_{i=1}^{n}\left(\dfrac{1}{n-1}\sum\limits_{j\neq i}x_{j}\right)\\ &=\sum\limits_{i=1}^{n}f\left(\dfrac{1-x_{i}}{n-1}\right). \end{aligned}
于是引理成立.

回证原题.令f\left(x\right)=\ln\dfrac{1+x}{x},注意到,对任意x,y\in \left(0,1\right),都有
\begin{aligned} f(x)+f(y) &=\ln \dfrac{1+x}{x}+\ln \dfrac{1+y}{y}\\ &=\ln \dfrac{1+x y+x+y}{x y} \\ &=\ln \left(\dfrac{1}{x y}+\dfrac{x+y}{x y}+1\right) \\ & \geqslant \ln \left( \dfrac{1}{\left(\dfrac{x+y}{2}\right)^{2}}+\dfrac{x+y}{\left(\dfrac{x+y}{2}\right)^{2}}+1\right) \\ &=\ln \left(\dfrac{4}{(x+y)^{2}}+\dfrac{4}{x+y}+1\right)\\ &=\ln \left(\dfrac{(x+y+2)^{2}}{(x+y)^{2}}\right) \\ &=2 \ln \left(1+\dfrac{1}{\dfrac{x+y}{2}}\right)\\ &=2 f\left(\dfrac{x+y}{2}\right) \end{aligned}
所以,是\left(0,1\right)上的凸函数,依此结合前面所得可知命题成立.

2022-02-14-03

(来源: 数学奥林匹克小丛书 第二版 高中卷 数列与数学归纳法 冯志刚 习题二 P095 习题09)

斐波那契数列\left\{F_{n}\right\}满足:F_{1}=F_{2}=1,F_{n+2}=F_{n+1}+F_{n}.证明:\sum\limits_{i=1}^{n}\dfrac{F_{i}}{2^{i}}<2.

证明

S_{n}=\sum\limits_{i=1}^{n}\dfrac{F_{i}}{2^{i}},则S_{1}=\dfrac{1}{2},S_{2}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4},而当n\geqslant 3时,有
\begin{aligned} S_{n} &=\dfrac{1}{2}+\dfrac{1}{4}+\sum\limits_{i=3}^{n} \dfrac{F_{i}}{2^{i}} \\ &=\dfrac{3}{4}+\sum\limits_{i=3}^{n} \dfrac{F_{i-1}+F_{i-2}}{2^{i}} \\ &=\dfrac{3}{4}+\frac{1}{2} \sum\limits_{i=3}^{n} \dfrac{F_{i-1}}{2^{i-1}}+\dfrac{1}{4} \sum\limits_{i=3}^{n} \dfrac{F_{i-2}}{2^{i-2}} \\ &=\dfrac{3}{4}+\frac{1}{2} \sum\limits_{i=2}^{n-1} \dfrac{F_{i}}{2^{i}}+\dfrac{1}{4} \sum\limits_{i=1}^{n-2} \dfrac{F_{i}}{2^{i}} \\ &=\dfrac{3}{4}+\dfrac{1}{2}\left(S_{n-1}-\dfrac{1}{2}\right)+\frac{1}{4} S_{n-2} \\ &=\dfrac{1}{2}+\dfrac{1}{2} S_{n-1}+\dfrac{1}{4} S_{n-2} \end{aligned}
利用S_{1}=\dfrac{1}{2}S_{2}=\dfrac{3}{4}可知对n=1,2都有S_{n}<2;现设对n=k,k+1都有S_{n}<2,那么有
S_{k+2}=\dfrac{1}{2}+\dfrac{1}{2}S_{k+1}+\dfrac{1}{4}S_{k}<\dfrac{1}{2}+\dfrac{1}{2}\times 2+\dfrac{1}{4}\times 2=2.
所以,命题成立.

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