7.设$u_i\in L^{p_i}(\Omega),p_i>0,i=1,2,\cdots,m$,且${\sum\limits_{i = 1}^{m } {\frac{1}{{{p_i}}}} }=1$.证明 $$\iint\limits_\Omega {{u_1} \cdots {u_m}dxdy} \le {\left\| {{u_1}} \right\|_{{p_1}}} \cdots {\left\| {{u_m}} \right\|_{{p_m}}}.$$

解.利用P270的Holder不等式以及数学归纳法即可.假设$m-1$时不等式成立,则有

$$\begin{align*}&\iint\limits_\Omega {{u_1} \cdots {u_{m - 1}}{u_m}dxdy} \le \iint\limits_\Omega {\left| {{u_1} \cdots {u_{m - 1}}{u_m}} \right|dxdy} \\= &{\left\| {{u_1} \cdots {u_{m - 1}}{u_m}} \right\|_1} \le {\left\| {{u_1} \cdots {u_{m - 1}}} \right\|_{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}{\left\| {{u_m}} \right\|_{{p_m}}}\\\le& {\left( {{{\left\| {u_1^{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \right\|}_{{p_1}\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \cdots {{\left\| {u_{m - 1}^{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \right\|}_{{p_{m - 1}}\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}} \right)^{\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}{\left\| {{u_m}} \right\|_{{p_m}}}\\= &{\left\| {{u_1}} \right\|_{{p_1}}} \cdots {\left\| {{u_{m - 1}}} \right\|_{{p_{m - 1}}}}{\left\| {{u_m}} \right\|_{{p_m}}}.\end{align*}$$

另解:利用$n$元的Holder不等式

$$\prod\limits_{i = 1}^m {x_i^{{\theta _i}}} \le \sum\limits_{i = 1}^m {{\theta _i}{x_i}} ,\quad \text{其中}\sum\limits_{i = 1}^m {{\theta _i}} = 1,{\theta _i} \ge 0.$$

取 $${x_i} = \frac{{{{\left| {{u_i}} \right|}^{{p_i}}}}}{{\left\| {{u_i}} \right\|_{{p_i}}^{{p_i}}}},{\theta _i} = \frac{1}{{{p_i}}}$$ 再积分即可.

证明 $$1 < \iiint\limits_{{{\left[ {0,1} \right]}^3}} {\left( {\cos \left( {xyz} \right) + \sin \left( {xyz} \right)} \right)dxdydz} < \sqrt 2 .$$

证.注意到 $$1 < \cos \left( {xyz} \right) + \sin \left( {xyz} \right) = \sqrt 2 \sin \left( {xyz + \frac{\pi }{4}} \right) < \sqrt 2$$ 即可.

证明

$${\left\{ {{{\int_a^b {dx\left[ {\int_c^d {f\left( {x,y} \right)dy} } \right]} }^2}} \right\}^{1/2}} \le \int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} ,$$

其中$f$是连续函数.

证.利用Cauchy-Schwarz不等式我们有

$$\begin{align*}&{\left\{ {\int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} } \right\}^2}\\= &\int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} \cdot \int_c^d {dz{{\left[ {\int_a^b {{f^2}\left( {x,z} \right)dx} } \right]}^{1/2}}} \\\ge& \int_c^d {dy} \int_c^d {dz} \int_a^b {f\left( {x,y} \right)f\left( {x,z} \right)dx} = {\int_a^b {dx\left[ {\int_c^d {f\left( {x,y} \right)dy} } \right]} ^2}.\end{align*}$$

（13年武大数分）求$\displaystyle I = \iint\limits_\Sigma  {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{3}{2}}}{{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)}^{ - \frac{1}{2}}}dS}$,其中$\sum$为椭球面: $\displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1(a,b,c>0)$.

下面是自己的解答：

参考：http://www.math.org.cn/forum.php?mod=viewthread&tid=21850&extra=&page=2

前几天徐半仙问了我谢惠民下册P325页上一道难度稍大的全微分题目，利用今晚美好的独处时间（笑哭），做了下，解答如下：

对于以下一阶微分形式$\omega$,求函数$M(x,y)\neq0$, 使得在适当的区域内$M\omega$为全微分,并求其原函数:

（1） $\displaystyle \omega = \left[ { - y\sqrt {{x^2} + {y^2} + 1} - x\left( {{x^2} + {y^2}} \right)} \right]dx + \left[ {x\sqrt {{x^2} + {y^2} + 1} - y\left( {{x^2} + {y^2}} \right)} \right]dy$;

（2） $\displaystyle \omega = x\left[ {{{\left( {ay + bx} \right)}^3} + a{y^3}} \right]dx + y\left[ {{{\left( {ay + bx} \right)}^3} + b{x^3}} \right]dy$.

解：（1）取 $$M = \frac{1}{{\left( {{x^2} + {y^2}} \right)\sqrt {{x^2} + {y^2} + 1} }},$$ 我们有

$$M\omega = \left( { - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dx + \left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy.$$

则有$P = - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }},Q = \frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}$,且 $$\frac{{\partial P}}{{\partial y}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} + \frac{{xy}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} = \frac{{\partial Q}}{{\partial x}}.$$

此时原函数为

$$\begin{align*}\varphi \left( {x,y} \right) = &\int_{{x_0}}^x {P\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q\left( {x,y} \right)dy} + C'\\= &\int_{{x_0}}^x {\left( { - \frac{{{y_0}}}{{{x^2} + y_0^2}} - \frac{x}{{\sqrt {{x^2} + y_0^2 + 1} }}} \right)dx} + \int_{{y_0}}^y {\left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy} + C'\\= &\left( { - \arctan \frac{x}{{{y_0}}} - \sqrt {{x^2} + y_0^2 + 1} + \arctan \frac{{{x_0}}}{{{y_0}}} + \sqrt {x_0^2 + y_0^2 + 1} } \right)\\&+ \left( {\arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} - \arctan \frac{{{y_0}}}{x} + \sqrt {{x^2} + y_0^2 + 1} } \right) + C'\\=& \arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} + C.\end{align*}$$

值得一提的是：本题的积分因子是通过Wolfram Alpha求解出ODE，然后分别对$x,y$求偏导得来的.

（2）丁同仁书上一定理：

齐次方程$P(x,y)dx+Q(x,y)dy=0$有积分因子$M=\frac{1}{xP+yQ}$.

定理的证明:作变换$y=ux$,则由$P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0$是齐次方程,我们有 $$P\left( {x,ux} \right)dx + Q\left( {x,ux} \right)\left( {udx + xdu} \right) = \left[ {{x^m}P\left( {1,u} \right) + u{x^m}Q\left( {1,u} \right)} \right]dx + {x^{m + 1}}Q\left( {1,u} \right)du = 0.$$

方程两边同乘 $$\frac{1}{{xP + yQ}} = \frac{1}{{{x^{m + 1}}\left[ {P\left( {1,u} \right) + uQ\left( {1,u} \right)} \right]}},$$ 则有

$$\frac{1}{x}dx + \frac{{Q\left( {1,u} \right)}}{{P\left( {1,u} \right) + uQ\left( {1,u} \right)}}du = 0.$$ 显然此方程为全微分方程.证毕.

取 $$M = \frac{1}{{xP + yQ}} = \frac{1}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.$$

则有

$$P' = \frac{{x{{\left( {ay + bx} \right)}^3} + ax{y^3}}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}},Q' = \frac{{y{{\left( {ay + bx} \right)}^3} + b{x^3}y}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.$$

我猜此时一定成立 $$\frac{{\partial P'}}{{\partial y}} = \frac{{\partial Q'}}{{\partial x}}.$$

事实上

$$\begin{align*}\frac{{\partial P'}}{{\partial y}} = & - 2xy{\left( {ay + bx} \right)^6} - 2{x^3}y{\left( {ay + bx} \right)^4} + \left( {5a{x^3}{y^2} + ax{y^4}} \right){\left( {ay + bx} \right)^3}\\& - 3{a^2}x{y^3}\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + a{x^3}{y^4}\left( {ay + bx} \right) - {a^2}{x^3}{y^5}\\\frac{{\partial Q'}}{{\partial x}} = &- 2xy{\left( {ay + bx} \right)^6} - 2x{y^3}{\left( {ay + bx} \right)^4} + \left( {5b{x^2}{y^3} + b{x^4}y} \right){\left( {ay + bx} \right)^3}\\& - 3{b^2}{x^3}y\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + b{x^4}{y^3}\left( {ay + bx} \right) - {b^2}{x^5}{y^3}.\end{align*}$$

于是

$$\begin{align*}\varphi \left( {x,y} \right) &= \int_{{x_0}}^x {P'\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q'\left( {x,y} \right)dy} + C'\\&= \frac{1}{2}\ln \left[ {\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)} \right] - \frac{3}{2}\ln \left( {ay + bx} \right) + C.\end{align*}$$

事实上,我们还可取 $$M = \frac{1}{{{{\left( {ay + bx} \right)}^3}}},$$ 由此得到

$$\varphi \left( {x,y} \right) = \frac{{{x^2} + {y^2}}}{2} + \frac{{{x^2}{y^2}}}{{2{{\left( {ay + bx} \right)}^2}}} + C.$$

这道题来自MAA的杂志The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), pp. 500-507,可以参考链接http://www.jstor.org/stable/10.4169/amer.math.monthly.122.5.500?seq=1#page_scan_tab_contents

求 $$\int_0^\infty  {\frac{1}{x}dx} \int_0^x {\frac{{\cos \left( {x - y} \right) - \cos x}}{y}dy} .$$

解.(翻译而来)令$f\left( {x,y} \right) = \frac{{\cos \left( {x - y} \right) - \cos x}}{y}$.对$x>0$,我们有 $$\int_0^x {f\left( {x,y} \right)dy}  = \int_0^1 {\frac{{\cos \left( {1 - t} \right)x - \cos x}}{y}dt}  = x\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\, dudt} .$$

因而对$R>0$,

$$\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx}  = \int_0^R {\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\,dudtdx} } .$$

而$|\sin ux|\leq1$,该三重积分是绝对收敛的.由Fubini定理可知积分能交换次序

练习题1.证明： $$\sum\limits_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}\leq p,\,\,(p\ge1).$$

证:由Lagrange中值定理,我们有

$$\sqrt[p]{{n + 1}} - \sqrt[p]{n} = \frac{1}{p}{\xi ^{1/p - 1}} \ge \frac{1}{p}{\left( {n + 1} \right)^{1/p - 1}},\quad \xi  \in \left( {n,n + 1} \right).$$

因此 $$\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}} = \frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} \cdot \frac{{{{\left( {n + 1} \right)}^{1/p - 1}}}}{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}} \le p\frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} = p\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right).$$

立即有

$$\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}}}  \le p\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right)}  = p.$$

练习题2.设$\displaystyle S_n=\sum\limits_{k=1}^{n}a_k, p>1,c>1$,证明： $$\sum\limits_{n=1}^{\infty}\frac{S_n^p}{n^c}\le K\sum\limits_{n=1}^{\infty}\frac{(na_n)^p}{n^c},$$ 并求出$K$的最优值.

练习题3.设$a_n$是有界的正数列，$p>0$，证明：

证.注意到

$$\begin{align*}\sum\limits_{k = 1}^\infty  {{a_k}}  &= \sum\limits_{k = 1}^\infty  {\arctan \frac{1}{{{k^2} + k + 1}}}  = \sum\limits_{k = 1}^\infty  {\left( {\arctan \frac{1}{k} - \arctan \frac{1}{{k + 1}}} \right)}  = \frac{\pi }{4}\\\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^4}}}}  &= \zeta \left( 4 \right) = \frac{{{\pi ^4}}}{{90}}.\end{align*}$$

由Cauchy-Schwarz不等式可知

$$\sum\limits_{k = 1}^N {\frac{1}{{{k^4}}}}  \cdot \sum\limits_{k = 1}^N {{a_k}}  \ge {\left( {\sum\limits_{k = 1}^N {\frac{{a_k^{1/2}}}{{{k^2}}}} } \right)^2}.$$

令$N\to\infty$,我们有 $$\sum\limits_{k = 1}^\infty  {\frac{{a_k^{1/2}}}{{{k^2}}}}  \le \sqrt {\frac{{{\pi ^4}}}{{90}} \cdot \frac{\pi }{4}}  = \sqrt {\frac{{{\pi ^5}}}{{360}}}  < \sqrt {\frac{\pi }{3}} .$$

也可通过放缩实现 $$\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^4}}}}  = 1 + \sum\limits_{k = 2}^\infty  {\frac{1}{{{k^4}}}}  < 1 + \sum\limits_{k = 2}^\infty  {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \frac{4}{3}.$$

练习题10.设$\displaystyle a_n > 0, S_n=\sum\limits_{k=1}^na_k$，证明：

(2)如法炮制.由柯西不等式我们得

$$\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}} \ge {\left( {1 + 2 + \cdots + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},$$

即

$$\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}} \le \frac{{4\left( {2n + 1} \right)}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .$$

因此

$$\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}}} &\le 4\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} } = 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}} } \\&= 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} } = 4\sum\limits_{m = 1}^\infty {\frac{1}{{{a_m}}}}.\end{align*}$$

练习题11.设$\displaystyle a_n \ge 0, n=1,2,\ldots,\sum\limits_{n=1}^{\infty}a_n < \infty$,证明:

$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}a_n$$ ,且证明$e$是最优值.

此题再拓展下求证: $$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}[1-\frac{1}{2(n+1)}]a_n.$$

练习题12.如果正项级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{p_n}$收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{n^2}{(p_1+p_2+\cdots+p_n)^2}p_n$也收敛.

练习题13.设$\displaystyle \sum\limits_{n=1}^{\infty}a_n$为正项级数,且$\displaystyle \sum\limits_{k=1}^{n}(a_k-a_n)$对$n$有界,$a_n$单调递减趋于$0$,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛.

练习题14.设级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛, $\displaystyle \sum\limits_{n=1}^{\infty}(b_{n+1}-b_n)$绝对收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_nb_n$收敛.

练习题15.设$a_n>0,\left\{ a_n-a_{n+1}\right\}$为一个严格递减的数列.如果$\sum_{n=1}^{\infty}a_n$收敛。试证： $$\lim\limits_{n \to \infty}\left( \cfrac{1}{a_{n+1}}-\cfrac{1}{a_n}\right)=+\infty.$$

练习题16.能否构造一个收敛数列$\sum\limits_{n=1}^{\infty}a_n$,使得级数$\sum\limits_{n=1}^{\infty}a_n^3$发散.

练习题17.设$\lim \limits_{n\rightarrow +\infty}x_n=+\infty$,正项级数$\sum\limits_{n=1}^{\infty}y_n$收敛,设$n_0$是某一自然数,

练习题18.设$\sum\limits_{n=1}^{\infty}a_n$是一正项收敛级数,且有$a_{n+1}< \frac{1}{2}(a_n+a_{n+2}),\,\frac{1}{a_{n+2}}-\frac{1}{a_{n+1}}\le \frac{1}{3}(\frac{1}{a_{n+3}}-\frac{1}{a_{n}})$,

证明：对$n\geq 3$有 $$\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin{(2n+1)t}}{\sin{t}}\right|dt<\pi\left(1+\frac{\ln{n}}{2}\right).$$

Poof.For all $x$,

链接:http://math.stackexchange.com/questions/763253/how-prove-this-integral-inequality-6-left-int-01fxdx-right2-le-1-8-i?rq=1

逻辑丁的提问:证明 $$\sum\limits_{k = 1}^{+\infty} {\frac{{\sin kx}}{{{k^a}}}}  > 0,x \in \left( {0,\pi } \right),a \in \left( {0,\frac{1}{2}} \right]$$ 证明在$(0,\pi)$上勒贝格可积.

往事如烟！

谢上P9的一道不等式题，以前写过，但文件丢失，先前的解答难以回忆起，现在重新给出解答。

用向前-向后数学归纳法证明:设$\displaystyle 0<x_i\leq \frac12,i=1,2,\cdots,n$,则

$$\frac{{\prod\limits_{i = 1}^n {{x_i}} }}{{{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^n}}} \le \frac{{\prod\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} }}{{{{\left[ {\sum\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} } \right]}^n}}}.$$

(这个不等式是由在美国数学界有重大影响的华裔数学家樊畿(Fan Ky)得到的,关于它的许多研究和推广见[30].)

首先,

$$\frac{{{{\left[ {\sum\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} } \right]}^n}}}{{{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^n}}} \le \frac{{\prod\limits_{i = 1}^n {\left( {1 - {x_i}} \right)} }}{{\prod\limits_{i = 1}^n {{x_i}} }} \Leftrightarrow {\left[ {\frac{n}{{\sum\limits_{i = 1}^n {{x_i}} }} - 1} \right]^n} \le \prod\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}} - 1} \right)} .$$

当$n=2$时,即证

$$\left( {\frac{1}{{{x_1}}} - 1} \right)\left( {\frac{1}{{{x_2}}} - 1} \right) \ge {\left[ {\frac{2}{{{x_1} + {x_2}}} - 1} \right]^2}.$$

展开得

$$\frac{1}{{{x_1}{x_2}}} - \frac{1}{{{x_1}}} - \frac{1}{{{x_2}}} + 1 \ge \frac{4}{{{{\left( {{x_1} + {x_2}} \right)}^2}}} - \frac{4}{{{x_1} + {x_2}}} + 1.$$

等价于证明

$$\frac{1}{{{x_1}{x_2}}} - \frac{4}{{{{\left( {{x_1} + {x_2}} \right)}^2}}} \ge \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} - \frac{4}{{{x_1} + {x_2}}} \Leftrightarrow \frac{{{{\left( {{x_1} - {x_2}} \right)}^2}}}{{{{\left( {{x_1} + {x_2}} \right)}^2}{x_1}{x_2}}} \ge \frac{{{{\left( {{x_1} - {x_2}} \right)}^2}}}{{\left( {{x_1} + {x_2}} \right){x_1}{x_2}}}.$$

注意到$x_1+x_2\leq 1$,上式显然成立.

从$n=2$的已知情况出发,可以得到如下$n=4$时的情形:

$$\begin{align*}&\prod\limits_{i = 1}^4 {\left( {\frac{1}{{{x_i}}} - 1} \right)} = \prod\limits_{i = 1}^2 {\left( {\frac{1}{{{x_i}}} - 1} \right)} \cdot \prod\limits_{i = 3}^4 {\left( {\frac{1}{{{x_i}}} - 1} \right)} \ge {\left[ {\frac{2}{{\sum\limits_{i = 1}^2 {{x_i}} }} - 1} \right]^2}{\left[ {\frac{2}{{\sum\limits_{i = 3}^4 {{x_i}} }} - 1} \right]^2}\\= &{\left[ {\left( {\frac{1}{{\frac{1}{2}\sum\limits_{i = 1}^2 {{x_i}} }} - 1} \right)\left( {\frac{1}{{\frac{1}{2}\sum\limits_{i = 3}^4 {{x_i}} }} - 1} \right)} \right]^2} \le {\left[ {{{\left( {\frac{2}{{\frac{1}{2}\sum\limits_{i = 1}^2 {{x_i}} + \frac{1}{2}\sum\limits_{i = 3}^4 {{x_i}} }} - 1} \right)}^2}} \right]^2} = {\left[ {\frac{4}{{\sum\limits_{i = 1}^4 {{x_i}} }} - 1} \right]^4}.\end{align*}$$

同样可知,若$n=2^k$时不等式已成立,则可得到$n=2^{k+1}$时的不等式

$$\begin{align*}&\prod\limits_{i = 1}^{{2^{k + 1}}} {\left( {\frac{1}{{{x_i}}} - 1} \right)} = \prod\limits_{i = 1}^{{2^k}} {\left( {\frac{1}{{{x_i}}} - 1} \right)} \cdot \prod\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {\left( {\frac{1}{{{x_i}}} - 1} \right)} \ge {\left[ {\frac{{{2^k}}}{{\sum\limits_{i = 1}^{{2^k}} {{x_i}} }} - 1} \right]^{{2^k}}}{\left[ {\frac{{{2^k}}}{{\sum\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right]^{{2^k}}}\\= & {\left[ {\left( {\frac{1}{{\frac{1}{{{2^k}}}\sum\limits_{i = 1}^{{2^k}} {{x_i}} }} - 1} \right)\left( {\frac{1}{{\frac{1}{{{2^k}}}\sum\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right)} \right]^{{2^k}}} \le {\left[ {{{\left( {\frac{2}{{\frac{1}{{{2^k}}}\sum\limits_{i = 1}^{{2^k}} {{x_i}} + \frac{1}{{{2^k}}}\sum\limits_{i = {2^k} + 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right)}^2}} \right]^{{2^k}}} = {\left[ {\frac{{{2^{k + 1}}}}{{\sum\limits_{i = 1}^{{2^{k + 1}}} {{x_i}} }} - 1} \right]^{{2^{k + 1}}}}.\end{align*}$$

这样就证明了当$n$为$2$的所有方幂时平均值不等式已成立.这是“向前”部分.

第二步要证明,当平均值不等式对某个$n>2$成立时,则它对$n-1$也一定成立.这是证明中的“向后”部分.写出

$$\begin{align*}&{\left[ {\frac{{n - 1}}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^{n - 1}} = {\left[ {\frac{n}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} + \frac{1}{{n - 1}}\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^n} \cdot {\left[ {\frac{{n - 1}}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^{ - 1}}\\\le &\prod\limits_{i = 1}^{n - 1} {\left( {\frac{1}{{{x_i}}} - 1} \right)} \cdot \left( {\frac{1}{{\frac{1}{{n - 1}}\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right) \cdot {\left[ {\frac{{n - 1}}{{\sum\limits_{i = 1}^{n - 1} {{x_i}} }} - 1} \right]^{ - 1}} = \prod\limits_{i = 1}^{n - 1} {\left( {\frac{1}{{{x_i}}} - 1} \right)} .\end{align*}$$

于是$n-1$时不等式也成立.合并以上向前和向后两部分,可见不等式对每个自然数$n$成立.

事实上,我们有

$${\left[ {\frac{n}{{\sum\limits_{i = 1}^n {{x_i}} }} - 1} \right]^n} \le \prod\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}} - 1} \right)}  \Leftrightarrow \frac{1}{n}\sum\limits_{i = 1}^n {\ln \left( {\frac{1}{{{x_i}}} - 1} \right)}  \ge \ln \left( {\frac{1}{{\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} }} - 1} \right).$$ 结合函数$y = \ln \left( {\frac{1}{x} - 1} \right)$的凹凸性便可得证.