凌禹數(shù)學(xué)不等式問題100例

凌禹數(shù)學(xué)不等式問題100例

凌禹數(shù)學(xué)編輯部在2019年7月發(fā)布了一些符合高聯(lián)賽風(fēng)格的n元不等式題目，供大家練習(xí)。以下是其中的一部分題目：題1：給定正奇數(shù)??，??1,??2,…,????為??個非負(fù)實(shí)數(shù)。令????=????^2+????+1^2，????=2????????+1+1，其中??=1,2,…,??，且????+1=??1。定義??=min{??1,??2,…,????}，??=max{??1,??2,…,????}。證明：??≤??。題2：給定正整數(shù)??≥4，非負(fù)實(shí)數(shù)??1,??2,…,????滿足??1+??2+?+????=2。求??1+??2+?+????/[(??1^2+1)(??2^2+1)?(????^2+1)]的最小值。題3：給定整數(shù)??≥2，實(shí)數(shù)??1,??2,…,????≥?1且滿足??1+??2+?+????=0。求∑????????+1的最大值和最小值，其中??=1,2,…,??-1，且????+1=??1。題4：給定正整數(shù)??，正實(shí)數(shù)??1,??2,…,????，求(1+??1)(1+??1+??2)?(1+??1+??2+?+????)/√(??1??2?????)的最小值。題5：給定整數(shù)??≥2，正實(shí)數(shù)??2,??3,…,????滿足??2??3?????=1。證明：1/[??(1+??2)^2(1+??3)^3?(1+????)^??]>??(??-1)/(4??-1)。題6：給定正整數(shù)??，正實(shí)數(shù)??,??1,??2,…,????滿足∑??(????^2)=∑????∑1≤??<??≤??(????-????)。證明：??,記????=∑??(1/∑1≤??≤??(1+????))。則1/∑1≤??≤??(1+????)>??/(??+1)。題7：給定整數(shù)??≥2，正實(shí)數(shù)??1,??2,…,????滿足∑??????=1。證明：(∑√????)(∑√(1+????))/??^2≤(∑(1+????))/(√(??+1))。}，都有??12+??22+?+????2≥λ(??12+??22+?+????2).題18.給定正整數(shù)$n\geq2$，實(shí)數(shù)$a_1,a_2,\cdots,a_n$滿足$\sum\limits_{i=1}^na_i=0$且$\sum\limits_{i=1}^na_i^2=1$。求$\sum\limits_{i=1}^n|a_i|$的最大值和最小值；$max\{a_1,a_2,\cdots,a_n\}$的最大值和最小值。題19.給定正整數(shù)$n\geq3$，非零實(shí)數(shù)$x_1,x_2,\cdots,x_n$滿足$x_1+\dfrac{1}{2n}x_2x_1+\cdots+n=0$。求證：$$\dfrac{|x_1x_2+x_2x_3+\cdots+x_nx_1|}{n}\leq\sum\limits_{i=1}^n|x_i|(\max\{|x_i|\}-\min\{|x_i|\})$$題20.給定正整數(shù)$n$，實(shí)數(shù)$x_1,x_2,\cdots,x_n$滿足$\sum\limits_{i=1}^nx_i^2=1$。求證：對任意整數(shù)$k\geq2$，存在$n$個不全為零的整數(shù)$a_i$，$|a_i|\leqk-1,i=1,2,\cdots,n$，使得$$(k-1)\sqrt{n}\cdot|\sum\limits_{i=1}^na_ix_i|\leqk^{n-1}$$題21.給定整數(shù)$n\geq2$，設(shè)實(shí)數(shù)$a_1,a_2,\cdots,a_n\in[0,1]$。證明：$$\sum\limits_{i=1}^na_i-\sum\limits_{i=1}^na_ia_{i+1}\leq\dfrac{1}{2}+\dfrac{1}{2n}$$其中$a_{n+1}=a_1$。題22.給定整數(shù)$n\geq2$，設(shè)實(shí)數(shù)$a_1,a_2,\cdots,a_n\in(0,1)$。證明：$$\sqrt{n-1}\left(\dfrac{1}{\sqrt{1-a_1}}+\dfrac{1}{\sqrt{1-a_2}}+\cdots+\dfrac{1}{\sqrt{1-a_n}}\right)<\dfrac{a_1a_2\cdotsa_n}{\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}}$$題23.給定整數(shù)$n\geq2$，設(shè)實(shí)數(shù)$a_1,a_2,\cdots,a_n$是不全為零的非負(fù)實(shí)數(shù)。對$1\leqk\leqn$，記$$m_k=\max_{1\leql\leqk}\left\{\sum\limits_{i=l}^ka_i\right\}$$證明：對任意正實(shí)數(shù)$\alpha$，滿足$m_k>\alpha$的$k$少于$\dfrac{\sum\limits_{i=1}^na_i}{\alpha}$個。題24.求最大的正實(shí)數(shù)$\lambda$，使得對任意的正整數(shù)$n$以及任意正實(shí)數(shù)$a_1,a_2,\cdots,a_n$都有$$\dfrac{1}{1+\sum\limits_{i=1}^n\dfrac{1}{a_i}}\geq\dfrac{\lambda}{\sum\limits_{i=1}^n\sqrt{a_i}}$$題25.對于實(shí)數(shù)列$\{a_n\}$，定義數(shù)列$\{b_n\}$如下：$b_1=a_1,b_{n+1}=a_{n+1}-\sqrt{\sum\limits_{i=1}^na_i^2}$。求最小的實(shí)數(shù)$\lambda$，使得對任意實(shí)數(shù)列$\{a_n\}$，都有$$\sum\limits_{i=1}^nb_i^2\geq\lambda\sum\limits_{i=1}^na_i^2$$題目42：給定整數(shù)n≥2，正實(shí)數(shù)a1,a2,...an。證明：a1a2?an(1?(a1+a2+?+an))1≤nn+1.改寫：對于給定的整數(shù)n≥2和正實(shí)數(shù)a1,a2,...an，證明不等式a1a2?an(1?(a1+a2+?+an))1≤nn+1成立。題目43：給定整數(shù)n，正實(shí)數(shù)a1,a2,...an。證明：n?1∑(1?aiai+1)≥?n+2(∑ai)∏ai/(ai+1)（i=1,2,...,n），其中an+1=a1。改寫：對于給定的整數(shù)n和正實(shí)數(shù)a1,a2,...an，證明不等式n?1∑(1?aiai+1)≥?n+2(∑ai)∏ai/(ai+1)（i=1,2,...,n），其中an+1=a1成立。題目44：給定正整數(shù)n≥3，a,b為給定的實(shí)數(shù)。實(shí)數(shù)x,x1,x2,...,xn滿足x+x1+x2+?+xn=a，x+x1+x2+?+xn=b。求x的取值范圍。改寫：對于給定的正整數(shù)n≥3和實(shí)數(shù)a,b，假設(shè)實(shí)數(shù)x,x1,x2,...,xn滿足x+x1+x2+?+xn=a，x+x1+x2+?+xn=b。求x的取值范圍。題目45：設(shè)x1,x2,...,xn是任意實(shí)數(shù)。證明：x1+x2/1+x22+?+xn/1+x2n<√n.改寫：對于任意實(shí)數(shù)x1,x2,...,xn，證明不等式x1+x2/1+x22+?+xn/1+x2n<√n成立。題目46：設(shè)有2n個實(shí)數(shù)a1,a2,...a2n滿足條件∑i=1(ai+1?ai)=1。求(a(n+1)+a(n+2)+?+a2n)?(a1+a2+?+an)的最大值。改寫：設(shè)有2n個實(shí)數(shù)a1,a2,...a2n滿足條件∑i=1(ai+1?ai)=1。求(a(n+1)+a(n+2)+?+a2n)?(a1+a2+?+an)的最大值。題目47：設(shè)x1,x2,...,xn與a1,a2,...an(n≥2)是滿足條件：x1+x2+?+xn=0,|x1|+|x2|+?+|xn|=1,a1≥a2≥?≥an的兩組任意實(shí)數(shù)。為了使|a1x1+a2x2+?+anxn|≤A(a1?an)成立，求A的最小值。改寫：設(shè)x1,x2,...,xn與a1,a2,...an(n≥2)是滿足條件：x1+x2+?+xn=0,|x1|+|x2|+?+|xn|=1,a1≥a2≥?≥an的兩組任意實(shí)數(shù)。求使得|a1x1+a2x2+?+anxn|≤A(a1?an)成立的A的最小值。題目48：設(shè)函數(shù)f:N*→R滿足條件：對任意m,n∈N*，都有|f(m)+f(n)?f(m+n)|≤mn。求證：對任意的n∈N*，都有|f(k)/k?(f(n)/n+f(n?1)/(n?1)+?+f(1)/1)/n(n?1)/2|≤4/k（k=1,2,...,n）。改寫：設(shè)函數(shù)f:N*→R滿足條件：對任意m,n∈N*，都有|f(m)+f(n)?f(m+n)|≤mn。證明對任意的n∈N*，都有|f(k)/k?(f(n)/n+f(n?1)/(n?1)+?+f(1)/1)/n(n?1)/2|≤4/k（k=1,2,...,n）。題目49：設(shè)x1,x2,...,xn是非負(fù)實(shí)數(shù)，且滿足∑i=1nx2i+(∑i=1n∑j=i+1nxi*xj)=n(n+1)/2。求x1+x2+?+xn。改寫：設(shè)x1,x2,...,xn是非負(fù)實(shí)數(shù)，且滿足∑i=1nx2i+(∑i=1n∑j=i+1nxi*xj)=n(n+1)/2。求x1+x2+?+xn的值。題目50：對于實(shí)數(shù)$x_1\leqx_2\leq\cdots\leqx_n$，證明$\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}+\sum\limits_{i=1}^{n-1}\frac{x_i}{x_{i+1}}\geqx_1+x_2+\cdots+x_n$。題目51：對于正整數(shù)$a_1,a_2,\cdots,a_n$，設(shè)$s=a_1+a_2+\cdots+a_n$，且對于所有的$i\in\{1,2,\cdots,n\}$，有$\frac{(n-1)a_n}{a_i}+a_i\geq\frac{n-1}{n}s$。證明：$\frac{a_1}{s}+\frac{a_2}{s}+\cdots+\frac{a_n}{s}\leq1+\sqrt{\frac{1}{n-1}}$。題目52：對于正整數(shù)$n$和實(shí)數(shù)$x,y_1,y_2,\cdots,y_n$，滿足$\sum\limits_{i=1}^n\left(\frac{1}{2i}\right)^2=\frac{1}{n}\sum\limits_{i=1}^ny_i^2=1$，以及$\lambda_1,\lambda_2,\cdots,\lambda_n\in[0,1]$。證明：$\left|\sum\limits_{i=1}^n\lambda_i(x^2-y_i^2)\right|\leq\sqrt{1-\left(\sum\limits_{i=1}^n\lambda_ix_iy_i\right)^2}$。題目53：對于排列$P_1,P_2,\cdots,P_n$，證明$\frac{1}{n}\sum\limits_{i=1}^n\frac{1}{1+P_i}>\frac{1}{n+1}\left(\frac{1}{1+P_1}+\frac{1}{1+P_1P_2}+\cdots+\frac{1}{1+P_1P_2\cdotsP_n}\right)$。題目54：對于整數(shù)$n\geq2$，不全為零的實(shí)數(shù)$x_1,x_2,\cdots,x_n$，滿足$\sum\limits_{i=1}^nx_i=0$，且對于任意正實(shí)數(shù)$t$，至多存在一對$(i,j)$滿足$|x_i-x_j|\geqt$。證明：$\frac{1}{n}\sum\limits_{i=1}^nx_i^2<\frac{(\max\limits_{1\leqi\leqn}|x_i|-\min\limits_{1\leqi\leqn}|x_i|)^2}{4(n-1)}$。題目55：對于正整數(shù)$n$和正實(shí)數(shù)$c$，以及實(shí)數(shù)$x_1,x_2,\cdots,x_{2n}$，滿足$\sum\limits_{i=1}^{2n}x_i=c$，且對于任意$1\leqk\leq2n$，有$|x_{k+1}-x_k|<1$。證明：存在$n$個整數(shù)$i_1,i_2,\cdots,i_n$，滿足$\left|\sum\limits_{j=1}^nx_{i_j}-\frac{c}{n}\right|\leq\frac{k}{n}\sqrt{\frac{c}{n}}$。題目56：對于正整數(shù)$n\geq2$，定義常數(shù)$C(n)$為滿足對于任意$x_1,x_2,\cdots,x_n\in(0,1)$，且對于任意$1\leqi<j\leqn$，有$(1-x_i)(1-x_j)\geq\frac{1}{4}$，都有$\sum\limits_{i=1}^nx_i\geqC(n)\sum\limits_{1\leqi<j\leqn}\left(2x_ix_j+\sqrt{x_ix_j}\right)$。求$C(n)$的最大值。題85：給定實(shí)數(shù)序列$a_1,a_2,\dots,a_n$和$b_1,b_2,\dots,b_n$，滿足$\sum\limits_{i=1}^na_i^2=\sum\limits_{i=1}^nb_i^2=1$且$\sum\limits_{i=1}^na_ib_i=0$。證明：$\left(\sum\limits_{i=1}^na_i\right)^2+\left(\sum\limits_{i=1}^nb_i\right)^2\leqn$。證明：由柯西-施瓦茨不等式，有\(zhòng)begin{align*}\left(\sum_{i=1}^na_i\right)^2&=\left(\sum_{i=1}^na_i\cdot1\right)^2\\&\leq\left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^n1^2\right)\\&=n\sum_{i=1}^na_i^2\\&=n.\end{align*}同理，$\left(\sum\limits_{i=1}^nb_i\right)^2\leqn$。因此，$$\left(\sum_{i=1}^na_i\right)^2+\left(\sum_{i=1}^nb_i\right)^2\leq2n.$$又因?yàn)?\sum\limits_{i=1}^na_ib_i=0$，所以$$\left(\sum_{i=1}^na_i\right)^2+\left(\sum_{i=1}^nb_i\right)^2=n-2\sum_{i=1}^na_ib_i\leqn.$$綜上所述，得證。題86：給定非負(fù)實(shí)數(shù)$a_1,a_2,\dots,a_n$，證明：$$\sqrt[n]{\prod\limits_{i=1}^na_i}\leq\frac{1}{n}\sum\limits_{k=1}^n\frac{1}{\sqrt[k]{\prod\limits_{i=1}^na_i}}.$$證明：設(shè)$b_k=\sqrt[k]{\prod\limits_{i=1}^na_i}$，則$b_1,b_2,\dots,b_n$單調(diào)遞減。因此，\begin{align*}\sum_{k=1}^n\frac{1}{b_k}&=\frac{1}{b_1}+\frac{1}{b_2}+\cdots+\frac{1}{b_n}\\&=\frac{b_2b_3\cdotsb_n}{b_1b_2\cdotsb_n}+\frac{b_1b_3\cdotsb_n}{b_1b_2\cdotsb_n}+\cdots+\frac{b_1b_2\cdotsb_{n-1}}{b_1b_2\cdotsb_n}\\&=\frac{\sum\limits_{i=1}^na_i}{\sqrt[n]{\prod\limits_{i=1}^na_i}}\\&=\frac{1}{\sqrt[n]{\prod\limits_{i=1}^na_i}}.\end{align*}由均值不等式，得$$\frac{\sum\limits_{k=1}^n\frac{1}{b_k}}{n}\geq\sqrt[n]{\prod\limits_{i=1}^na_i},$$即$$\sqrt[n]{\prod\limits_{i=1}^na_i}\leq\frac{1}{n}\sum\limits_{k=1}^n\frac{1}{\sqrt[k]{\prod\limits_{i=1}^na_i}}.$$證畢。題87：給定正實(shí)數(shù)$x_1,x_2,\dots,x_n$，其中下標(biāo)按模$n$理解。證明：$$5x_i>\frac{1}{12}(x_{i+1}+x_{i+2})\quad(i=1,2,\dots,n).$$證明：由均值不等式，有$$\frac{x_{i+1}+x_{i+2}}{2}\geq\sqrt{x_{i+1}x_{i+2}},$$即$$\frac{1}{12}(x_{i+1}+x_{i+2})\geq\frac{1}{6}\sqrt{x_{i+1}x_{i+2}}.$$因此，\begin{align*}5x_i&>\frac{1}{6}\sqrt{x_{i+1}x_{i+2}}\\&\geq\frac{1}{6}\sqrt[3]{x_{i+1}^2x_{i+2}}\\&=\frac{1}{12}\sqrt[3]{x_{i+1}^2x_{i+2}\cdotx_{i+1}x_{i+2}^2}\\&=\frac{1}{12}\sqrt[3]{x_{i+1}^3x_{i+2}^3}\\&=\frac{1}{12}(x_{i+1}x_{i+2})^{\frac{3}{2}}\\&>\frac{1}{12}(x_{i+1}+x_{i+2}).\end{align*}證畢。題88：給定非負(fù)實(shí)數(shù)$x_1,x_2,\dots,x_n$，滿足$x_1+x_2+\cdots+x_n=1$，$0\leq\lambda\leqn$。證明：$$\sum_{i=1}^n\sqrt{x_i}\leq\sqrt{n}\sqrt[4]{\frac{n+\lambda}{\lambda}}\sqrt[4]{\sum_{i=1}^nx_i(1-x_i)}.$$證明：由柯西-施瓦茨不等式，有\(zhòng)begin{align*}\left(\sum_{i=1}^n\sqrt{x_i}\right)^2&\leq\left(\sum_{i=1}^nx_i\right)\left(\sum_{i=1}^n\frac{1}{\sqrt{x_i}}\right)\\&=1+\sum_{i=1}^n\frac{1}{\sqrt{x_i}}.\end{align*}因此，只需證明$$\sum_{i=1}^n\frac{1}{\sqrt{x_i}}\leq\sqrt{n}\sqrt[4]{\frac{n+\lambda}{\lambda}}\sqrt[4]{\sum_{i=1}^nx_i(1-x_i)}-1.$$設(shè)$f(x)=\dfrac{1}{\sqrt{x}}$，則$f''(x)=\dfrac{3}{4}x^{-\frac{5}{2}}>0$，即$f(x)$為凸函數(shù)。因此，由琴生不等式，得$$\sum_{i=1}^n\frac{1}{\sqrt{x_i}}\leqn\cdot\frac{1}{\sqrt{\frac{1}{n}\sum\limits_{i=1}^nx_i}}=\sqrt{n}\sqrt{\sum_{i=1}^nx_i}.$$因?yàn)?x_1+x_2+\cdots+x_n=1$，所以$\sum\limits_{i=1}^nx_i(1-x_i)=\sum\limits_{i=1}^nx_i-\sum\limits_{i=1}^nx_i^2\geq\dfrac{1}{n}-\dfrac{1}{n^2}=\dfrac{n-1}{n^2}$。因此，$$\sqrt{n}\sqrt{\sum_{i=1}^nx_i}\leq\sqrt{n}\sqrt[4]{\frac{n+\lambda}{\lambda}}\sqrt[4]{\sum_{i=1}^nx_i(1-x_i)}.$$綜上所述，得證。題89：給定非負(fù)實(shí)數(shù)$a_1,a_2,\dots,a_{100}$，滿足$\sum\limits_{i=1}^{100}a_i^2=1$。證明：$$\sum_{i=1}^{50}a_{2i-1}a_{2i}\leq\frac{1}{2}.$$證明：由柯西-施瓦茨不等式，有$$\sum_{i=1}^{50}a_{2i-1}a_{2i}\leq\sqrt{\left(\sum_{i=1}^{50}a_{2i-1}^2\right)\left(\sum_{i=1}^{50}a_{2i}^2\right)}=\frac{1}{2}.$$證畢。題90：給定正實(shí)數(shù)$x_1,x_2,\dots,x_n$，滿足$x_1+x_2+\cdots+x_n=1$。證明：$$\sum_{i=1}^n\sqrt{x_i^2+x_{i+1}^2}\leq2-\frac{1}{n}.$$證明：由柯西-施瓦茨不等式，有\(zhòng)begin{align*}\left(\sum_{i=1}^n\sqrt{x_i^2+x_{i+1}^2}\right)^2&\leq\left(\sum_{i=1}^n(x_i+x_{i+1})\right)\left(\sum_{i=1}^n\frac{x_i^2+x_{i+1}^2}{x_i+x_{i+1}}\right)\\&=2\sum_{i=1}^n(x_i^2+x_{i+1}^2)\\&=2\left(\sum_{i=1}^nx_i^2+2\sum_{i=1}^{n-1}x_ix_{i+1}\right)\\&=2\left(\sum_{i=1}^nx_i^2+2\sum_{i=1}^{n}x_ix_{i+1}\right)-4\sum_{i=1}^nx_i^2\\&=2\left(\sum_{i=1}^nx_i^2+\sum_{i=1}^{n}(x_i+x_{i+1})^2\right)-4\sum_{i=1}^nx_i^2\\&=2\left(\sum_{i=1}^nx_i^2+\sum_{i=1}^{n}x_i^2+2\sum_{i=1}^{n-1}x_ix_{i+1}+\sum_{i=1}^{n-1}x_{i+1}^2\right)-4\sum_{i=1}^nx_i^2\\&=2\left(\sum_{i=1}^nx_i^2+\sum_{i=1}^{n}x_i^2+2\sum_{i=1}^{n-1}x_ix_