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Author |
hbghlyj
Posted at 2020-1-17 13:25:31
第七组.
1.给定整数n$\geq2$,实数$x_1,x_2,\cdots,x_n\in[0,1]$,证明：$\sum\limits_{1\leq k < j\leq n}kx_kx_j\leq\frac{n-1}3\sum\limits_{k=1}^nkx_k$
法①当n=2时,ab$\le\frac13a+\frac23b\Leftrightarrow\left( {a - \frac{1}{3}} \right)\left( {b - \frac{2}{3}} \right) \leqslant \frac{1}{3} \bullet \frac{2}{3}$,当且仅当a=1,b=0时取等.
$kx_kx_j\le\frac k3x_k+\frac{2k}3x_j$,两边对$1\le k<j\le n$求和得???
法②由于两边关于$x_k$是一次式,取最值时$x_k$=0或1,设$x_{a_1}=x_{a_2}=\cdots=x_{a_m}=1$,其余$x_i$为0.
$(m-1)a_1+(m-2)a_2+\cdots+a_{m-1}\le\frac{n-1}3(a_1+a_2+\cdots+a_n)$.设m-i-1$\le\frac{n-1}3\le m-i,\therefore$可认为$a_1,a_2,\cdots,a_m$为连续m个正整数,???
2.$n\in\mathbb N_+,a_1,a_2,\cdots,a_n\in\mathbb R,r_1,r_2,\cdots,r_n\in\mathbb R_+,$证明：$\sum\limits_{i=1}^n\sum\limits_{j=1}^na_ia_j\min(r_i,r_j)\geq0$
由于对任意i,j,交换$a_i,a_j$不等式不变,不妨设$r_1\ge r_2\ge \cdots\ge r_n,$左式=$r_1a_1^2+r_2(2a_1a_2+a_2^2)+r_3(2a_1a_3+2a_2a_3+a_3^2)+\cdots=(r_1-r_2)a_1^2+(r_2-r_3)(a_1^2+2a_1a_2+a_2^2)+(r_3-r_4)(a_1^2+2a_1a_2+a_2^2+2a_1a_3+2a_2a_3+a_3^2)+\cdots+(r_{n-1}-r_n)(a_1^2+\cdots+a_{n-1}^2)+r_n(a_1^2+\cdots+a_n^2)\geq0$
令$f_i(x)=\led
{a_i}\quad 0 \le x \le r \\
0\quad \text{otherwise}
\endled$
$f_i(x) \bullet f _j(x) =\led \begin{array}{l}
{a_i}{a_j}\end{array}\quad 0 \le x \le \min \{ {r_i},{r_j}\} \\
0\quad \text{otherwise}\endled$
$F(x) = \mathop \smallint \nolimits_0^{ + \infty } {f_i}(x){f_j}(x){\rm{d}}x = {f_1}(x) + {f_2}(x) + \cdots + {f_n}(x) = \mathop \smallint \nolimits_0^{ + \infty } {(F(x))^2}{\rm{d}}x > 0$
3.切比雪夫不等式的加强：若一个实数列满足:从第二项起,每项均不大于前面所有项的算术平均值,则称这个实数列是伪减的.已知两个数列$\{a_n\},\{b_n\}$是伪减的,则$\sum\limits_{i=1}^na_ib_i\geq\frac1n\sum\limits_{i=1}^na_i\sum\limits_{i=1}^nb_i$
对n归纳，n=1时为恒等式，n=2时.设对n成立,下证对n+1成立:
$\sum\limits_{i = 1}^{n + 1} {{a_i}} {b_i} \ge \frac{1}{n}\sum\limits_{i = 1}^n {{a_i}} \sum\limits_{i = 1}^n {{b_i}} + {a_{n + 1}}{b_{n + 1}}$.
4.由正整数组成的有限集A满足：A的任意两个不同的子集的所有元素之和均不同,证明：A的所有元素的倒数之和小于2
5.设整数n$\geq$3,$a_1,a_2,\cdots,a_n$是实数,S是{1,2$\cdots$n}的一个非空子集,证明$\sum\limits_{i\in S}\leq\sum\limits_{1\leq i\leq j\leq n}(a_i+\cdots+a_j)^2$
6.设$x_1,x_2,\cdots,x_n\in\mathbf R$,证明:$\sum\limits_{i=1}^n\sum\limits_{j=1}^n\abs{x_i+x_j}\geq n\sum\limits_{i=1}^n\abs{x_i}$
7.设整数n$\geq$2,正实数$x_1,x_2\cdots x_n,y_1,y_2\cdots y_n$满足$x_1x_2\cdots x_n=y_1y_2\cdots y_n=1$,且对任意$1\leq j\leq n$都有$(y_i-1)(y_i-x_i)\leq0$.证明$x_1+x_2+\cdots+x_n\geq y_1+y_2+\cdots+y_n$
8.设整数n$\geq$4,正实数$x_1,x_2,\cdots,x_n$之积为1,证明$\sum\limits_{i=1}^n\frac{x_i^2}{(x_i+1)^2}\geq1$
9.给定整数$n\geq2,$求最小的常数d,使得不等式$\sum\limits_{i=1}^n\frac1{a_i^2+d}\leq\frac n{1+d}$对任意和为n的非负实数$a_1,a_2,\cdots,a_n$均成立.
10.设$a_1,a_2,\cdots,a_{10}$和$b_1,b_2,\cdots,b_{10}$是两个1,2,$\cdots$,10的排列,将$a_1b_1,a_2b_2,\cdots,a_{10}b_{10}$从大到小排成一列$M_1\geq M_2\geq\cdots\geq M_{10}$,求$M_1,M_2$的最值
11.设$a_1,a_2,\cdots,a_n$是一些实数,证明
(1)存在$\varepsilon_1,\varepsilon_2\cdots\varepsilon_n\in\{-1,1\}$,使得$\sum\limits_{i=1}^n\varepsilon_ia_i\leq\max\limits_{1\leq i\leq n}\abs{a_i}$
(2)存在不全为0的$\varepsilon_1,\varepsilon_2\cdots\varepsilon_n\in\{-1,0,1\}$,使得$\sum\limits_{i=1}^n\varepsilon_ia_i\leq\frac{\sum\limits_{i=1}^n\abs{a_i}}{2^n-1}$
12.设整数n$\geq3$,非负实数$a_1\geq a_2\geq\cdots\geq a_n$,将它们排在一个圆周上,证明
(1)对任意$1\leq i\leq\frac n2$,圆周上存在相邻两数之积不小于$a_ia_{n-i}$
(2)存在一种排列方式,使得圆周上任意相邻两数之积不大于$\max\limits_{1\leq i\le \frac n2}a_ia_{n-i}$
13.给定整数$n\geq3$,求最大的整数K=K(n),使得对任意n元实数组$(x_1,x_2,\cdots,x_n)$只要其任一排列$y_1,y_2,\cdots,y_n$满足$\sum\limits_{i=1}^{n-1}y_iy_{i+1}\geq-1$,就有$\sum\limits_{1\leq i\le j\leq n}x_ix_j\geq K$.
14.给定整数$n\geq5$,实数$x_1,x_2\cdots x_n$满足$-1\leq x_i\leq 1$(i=1,2$\cdots$n)及$\sum\limits_{i=1}^nx_i=0$.对$x_1,x_2\cdots x_n$的任一排列$y_1,y_2,\cdots,y_n$,计算$y_1y_2+y_2y_3+\cdots+y_{n-1}y_n+y_ny_1$的值,并设这些值中最大的一个为A,求A的最小可能值
15.设正整数$a_1,a_2,\cdots,a_{31},b_1,b_2,\cdots,b_{31}$满足
(1)$1\leq a_1\le a_2\le\cdots\le2015,1\leq b_1\le b_2\le\cdots\le b_{31}\leq 2015$,
(2)$a_1+a_2+\cdots+a_{31}=b_1+b_2+\cdots+b_{31}$,
求$\sum\limits_{i=1}^n\abs{a_i-b_i}$的最大可能值
16.$n\in\mathbf N_+,a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n,A,B\in\mathbf N_+,a_i\le b_i,a_i\le A,i=1,2,\cdots,n,\frac{b_1b_2\cdots b_n}{a_1a_2\cdots a_n}\leq\frac BA$,证明$\frac{(b_1+1)(b_2+1)\cdots(b_n+1)}{(a_1+1)(a_2+1)\cdots(a_n+1)}\le\frac{B+1}{A+1}$
$k_i=\frac{b_i}{a_i}\ge1,i=1,2,\cdots,n,$记$\frac BA=K$,条件化为$k_1k_2\cdots k_n\le K,$要证$\prod\limits_{i=1}^n\frac{k_ia_i+1}{a_i+1}\le\frac{KA+1}{A+1}.$对i=$1,2,\cdots,n,$由于$k_i\ge1$及$0<a_i\leq A$知,$\frac{k_ia_i+1}{a_i+1}=k_i-\frac{k_i-1}{a_i+1}\leq k_i-\frac{k_i-1}{A+1}=\frac{k_iA+1}{A+1}$.结合$K\ge k_1k_2\cdots k_n$知,只需证$A>0,k_i\ge1(i=1,2,\cdots,n)$时,有$\prod\limits_{i=1}^n\frac{k_iA+1}{A+1}\le\frac{k_1k_2\cdots k_nA+1}{A+1}.$
对n进行归纳.当n=1时,结论显然成立.当n=2时,由A>0,$k_1,k_2\ge1$可知$\frac{k_iA+1}{A+1}=\left(\prod\limits_{i=1}^m\frac{k_iA+1}{A+1}\right)\le\frac{k_1k_2\cdots k_mA+1}{A+1}\bullet\frac{k_{m+1}A+1}{A+1},$
17.不存在f:$\mathbf R^+\to\mathbf R^+$,使得$\left(f(x)\right)^2\ge f(x+y)\left(f(x)+y\right)$对任意x,y>0成立
18.对于固定的正整数s,t(s<t),设$a_n=\sum\limits_{i = s}^t {C_n^i} $,证明：对于所有正整数n,均有$a_n^2 \geqslant {a_{n - 1}}{{\text{a}}_{n + 1}}$
19.a,b,c$\in[0,1],a+b,b+c,c+a\ge1,$求证：$1-\frac{2\sqrt2abc}{\sqrt{a^2+b^2+c^2}}\leq(1-a)^2+(1-b)^2+(1-c)^2\leq1.$
20.设$a_1\geq a_2\geq\cdots\ge a_n>0$,给定k$\in\{1,2,\cdots,n\},$证明：可将$a_1,a_2,\cdots,a_n$分成k组,使得每组的所有数之和均不小于$\frac12\mathop {\min }\limits_{1 \leqslant j \leqslant k} \frac{{\sum\limits_{i = j}^n {{a_i}} }}{{k + 1 - j}}$
21.求所有正整数n>1,使得存在两个不是整数的有理数a,b满足$a+b,a^n+b^n\in\mathbf Z$
n为奇数,取$a=\frac13,b=\frac{3^n-1}3$满足条件;下证n为偶数时不存在这样的a,b:
$a=\frac xy,b=\frac zw,(x,y)=(z,w)=1,y,w>1,a+b\in\mathbf Z\Rightarrow y=w\left(\frac{xw+yz}{yw}\in\mathbf Z\Rightarrow y\mid xw\Rightarrow y\mid w\Rightarrow w\mid yz\Rightarrow w\mid y\Rightarrow y=w\right)$.
$z=ky-x,y^n\mid x^n+(ky-x)^n,x^n+(ky-x)^n=My^2-nkyx+2x^n\Rightarrow y\mid 2x^n\Rightarrow y|2\Rightarrow y=2\Rightarrow 2|x+1$，而$4|2x^n$，矛盾
22.对任给正整数m,n,总存在正整数k,使得$2^k-m$至少有n个不同的素因子
对n归纳,只需考虑m为奇数的情况.n=1时,$k_1=3m,2^{3m}-m>1$.假设n时成立,存在$k_n$使$A_n=2^{k_n}-m$至少含n个不同的素因子.当n+1时,$2^{k_n+\phi(A_n^2)}-m\equiv2^{k_n}-m\equiv A_n\pmod{A_n^2}$,令$k_{n+1}=k_n+\phi(A_n^2),A_n\mid2^{k_{n+1}}-m,A_n\not\mid\frac{2^{k_{n+1}-m}}{A_n},\frac{2^{k_{n+1}-m}}{A_n}\equiv1\pmod{A_n},$取$\frac{2^{k_{n+1}-m}}{A_n}$的质因数p,p与$A_n$中
23.非负数a,b,c,d和为100,求\[S = \sqrt[3]{{\frac{a}{{b + 7}}}} + \sqrt[3]{{\frac{{\text{b}}}{{c + 7}}}} + \sqrt[3]{{\frac{c}{{d + 7}}}} + \sqrt[3]{{\frac{d}{{a + 7}}}}\]的最大可能值
24.设m,n$\ge2$是整数,$f(x_1,x_2,\cdots,x_n)$是一个实系数多项式,$f(x_1,x_2,\cdots,x_n)=\lfloor\frac{x_1+x_2+\cdots+x_n}n\rfloor$对任意$x_1,x_2,\cdots,x_n\in\{0,1,\cdots,m-1\}$均成立.证明$\deg f\geq n$
25.互异正整数m,n,p,q满足m+n=p+q,$\sqrt m+\sqrt[3]n=\sqrt p+\sqrt[3]q\ge2004$，求[m,n,p,q]的最小值
令$m=a^2,n=b^3,p=c^2,q=d^3$,则$a+b=c+d\Rightarrow a-c=d-b,a^2+b^3=c^2+d^3\Leftrightarrow(a+c)(a-c)=(d-b)(d^2+bd+d^2)\Rightarrow a+c=d^2+bd+d^2$,令$a-c=d-b=1,b>2004\Rightarrow a=\frac{b^2+bd+d^2+1}2,c=\frac{b^2+bd+d^2-1}2\Rightarrow a,c\in\mathbf Z$
a=1954,b=62,c=2015,d=1,m=3818116,n=238328,p=4060225,q=1,[m,n,p,q]=961149422238200
这好像不能说明最小解的最小公倍数是最小的
26.$\sigma(n)$表示正整数n的正约数之和,证明：任给$\varepsilon>0$,存在无穷多个正整数n使$\frac{f(n)}n>\varepsilon$
令n=$k!$,$f(n)>n\left(1+\frac12+\cdots+\frac1k\right)$,$\frac{f(n)}n>1+\frac12+\cdots+\frac1k$,当$k\to\infty$时$RHS\to\infty$.
$任给$\varepsilon>0$,是否存在无穷多个正整数n使$\frac{f(n)}{n^2}>\varepsilon$
27.设m为正整数,素数p>m,证明：使得$m^2+n^2+p^2-2mn-2np-2pm$为完全平方数的正整数n的个数与p无关.
$m^2+n^2+p^2-2mn-2np-2pm=(m+n-p)^2-4mn$为完全平方数,$mx^2-(m+n-p)x+n$可分解为$(d_1x+a)(d_2x+b),d_1,d_2\in\mathbf N_+,d_1d_2=m,ab=n,-(ad_2+bd_1)=m+n-p,p=d_1d_2+ab-ad_2-bd_1=(d_1-a)(d_2-b),\{d_1-a,d_2-b\}=\{1,p\}$或$\{-1,-p\}$,$d_1+d_2-a-b=1+p>1+m=1+d_1d_2\ge d_1+d_2,d_1+d_2-a-b=-(1+p)<-(1+m)=1+d_1d_2\ge d_1+d_2$
$(m+n-p)^2-4mn=(2m+d_1p+d_2)^2-4m(m+p+d_1p+d_2)=4m^2+4m(d_1p+d_2)-4m^2-4m(p+d_1p+d_2)+(d_1p+d_2)^2=(d_1p-d_2)^2$
28.设m与n互素，证明：$m^{φ(n)}+n^{φ(m)}≡1\pmod{mn}$
由欧拉定理，$m^{φ(n)}+n^{φ(m)}≡n^{φ(m)}≡1\pmod{m}$，同理$m^{φ(n)}+n^{φ(m)}≡1\pmod{n}$，故得证.
29.设0≤a,b,c≤1, 求证$\sum\frac a{1+bc}≤2$
1.0≤u,v≤1=>0≤(1-u)(1-v)=>u+v-1≤uv
2.a+b+c≤2,a/(bc+1)+b/(ac+1)+c/(ab+1)≤a+b+c≤2
3.a+b+c-2≤(b+c-1)a≤abc.
ⅱ)a+b+c=a/(bc+1)+b/(ac+1)+c/(ab+1)+abc[1/(bc+1)+1/(ac+1)+1/(ab+1)]
≥a/(bc+1)+b/(ac+1)+c/(ab+1)+
[a+b+c-2][1/(1+1)+1/(1+1)+1/(1+1)]=
=a/(bc+1)+b/(ac+1)+c/(ab+1)+3/2[a+b+c-2]
==>
a/(bc+1)+b/(ac+1)+c/(ab+1)≤
a+b+c-3/2(a+b+c-2)=2-1/2(a+b+c-2)<2.
30.设0≤a,b,c≤1,证明:$\sum a{b+c+1}＋(1-a)(1-b)(1-c)≤1$
31.若$x^3+y^3=2$，求证:x+y≤2
法①假设x+y>2,x³+y³ > x³+(2-x)³ = 6x²-12x+8 = 6(x-1)²+2 ≥2
法②$\because x^3+y^3=(x+y)(x^2-xy+y^2)>0,x^2-xy+y^2>0,\therefore x+y>0,\because(x+y)(x^3+y^3)=x^4 + y^4 + x^3y + xy^3≥x^4 + y^4 + 2x^2y^2=(x^2+y^2)^2≥\frac14(x+y)^4,\therefore x+y\le 2$
32.$x_{ij}>0,\frac1{\sum\limits_{i=1}^m\frac1{\sum\limits_{j=1}^nx_{ij}}
}\ge\sum\limits_{j=1}^n\frac1{\sum\limits_{i=1}^m\frac1{x_{ij}}}$
这个不等式的物理意义是添加导线后电阻不增加
先证m=2的情形，即$\frac1{\frac1{\sum\limits_{j=1}^nx_{1j}}+\frac1{{\sum\limits_{j=1}^nx_{2j}}}}\ge\sum\limits_{j=1}^n\frac1{\frac1{x_{1j}}+\frac1{x_{2j}}}$
$\Leftrightarrow\frac{\sum\limits_{j=1}^nx_{1j}{\sum\limits_{j=1}^nx_{2j}}}{\sum\limits_{j=1}^nx_{1j}+{\sum\limits_{j=1}^nx_{2j}}}\ge\sum\limits_{j=1}^n\frac{x_{1j}x_{2j}}{x_{1j}+x_{2j}}$
$\Leftrightarrow\sum\limits_{j=1}^nx_{1j}-\frac{\left(\sum\limits_{j=1}^nx_{1j}\right)^2}{\sum\limits_{j=1}^nx_{1j}+{\sum\limits_{j=1}^nx_{2j}}}\ge\sum\limits_{j=1}^nx_{1j}-\sum\limits_{j=1}^n\frac{x_{1j}^2}{x_{1j}+x_{2j}}$
$\Leftrightarrow \frac{\left(\sum\limits_{j=1}^nx_{1j}\right)^2}{\sum\limits_{j=1}^nx_{1j}+{\sum\limits_{j=1}^nx_{2j}}}\le\sum\limits_{j=1}^n\frac{x_{1j}^2}{x_{1j}+x_{2j}}$
此即柯西不等式，当且仅当$x_{1j}:x_{2j}=\text{const}$时等号成立
假设命题对m为真，
$\frac1{\sum\limits_{i=1}^{m+1}\frac1{\sum\limits_{j=1}^nx_{ij}}
}\ge\frac1{\frac1{\sum\limits_{j=1}^n\frac1{\sum\limits_{i=1}^m\frac1{x_{ij}}}}+\frac1{\sum\limits_{j=1}^nx_{m+1,j}}
}\ge\sum\limits_{j=1}^n\frac1{\sum\limits_{i=1}^{m+1}\frac1{x_{ij}}}$(第二个$\ge$是根据m=2时的不等式)，因此命题对m+1也为真，由归纳定理得证。当且仅当$x_{1j}:x_{2j}:\cdots:x_{mj}=\text{const}$时等号成立。 |
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kuing
Posted at 2019-12-23 00:29:34
facebooker
Posted at 2019-12-23 17:36:10
