复旦大学国内外优秀数学教材推荐，参见http://www.library.fudan.edu.cn/wjzx/list/373-1-20.htm

证明\[\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.\]

解.(r9m)首先有

\begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4\left(1+\frac{2}{3n^2}\right)\left(1-\frac{1}{9n^2}\right)}\right).\end{align*}

分母里的无穷乘积$\displaystyle \prod\limits_{n=1}^{\infty}\left(1+\frac{2}{3n^2}\right)$和$\displaystyle \prod\limits_{n=1}^{\infty}\left(1-\frac{1}{9n^2}\right)$ 可以被忽略,当它们收敛于实数时.

因此

\begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right).\end{align*}

对分子进行分解

$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1).$$

我们得

$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}.$$

在$\displaystyle z = i,\frac{i}{3},\frac{i+1}{3},\frac{i-1}{3}$ 处运用$\displaystyle \frac{1}{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}$ .

我们可以改写为

$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}.$$

另一方面运用Gauss-Legendre Triplication Formula

$$ \Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right).$$

令$z = \dfrac{i-1}{3}$我们有

$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$

因此$$S = \arg \frac{-3^{i}\Gamma(i)}{\Gamma(i-1)} = \arg (3^{i}(1-i)) = \log 3 - \frac{\pi}{4}.$$

解法二.(robjohn)运用$\arctan(x)=\arg(1+ix)$,分解可知

\begin{align*}&1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}\\=&\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}.\end{align*}

因此有

\begin{align*}&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right).\end{align*}

裂项可知

\begin{align*}&\sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\lim_{m\to\infty}\sum_{n=1}^m\left[\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)\right]\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\arctan\left(\frac1n\right)\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\left[\frac1n+O\left(\frac1{n^3}\right)\right]\\=&\log(3)-\frac\pi4.\end{align*}

谢惠民下册P238第21章的一个参考题:

证明与曲面$ax^2+by^2+cz^2=1(abc\neq0)$相切的三个互相垂直的平面的交点在球面$x^2+y^2+z^2=\frac1a+\frac1b+\frac1c$上.

证:(幸子)椭球面方程为$ax^2+by^2+cz^2=1(abc\neq0)$,则法向量${n_i} = \left( {a{x_i},b{y_i},c{z_i}} \right)$,切平面方程为$a{x_i}x + b{y_i}y + c{z_i}z = 1$.

设三个切点分别为${\alpha _i}\left( {{x_i},{y_i},{z_i}} \right)\left( {i = 1,2,3} \right)$,三平面交点为$(x,y,z)$.由三平面垂直可知

\[\overrightarrow {{n_i}} \cdot \overrightarrow {{n_j}} = {a^2}{x_i}{x_j} + {b^2}{y_i}{y_j} + {c^2}{z_i}{z_j} = 0\left( {i \ne j} \right).\]

原点到三切平面的距离分别为

\[\frac{1}{{\sqrt {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} }}\left( {i = 1,2,3} \right).\]

由几何关系(考虑长方体对角线)可知

\[{x^2} + {y^2} + {z^2} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} .\]

设$\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)$,则对任意的$i = 1,2,3$,有

\[1 = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a{x_i}}&{b{y_i}}&{c{z_i}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right) = {t_i}\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right).\]

并且

\begin{align*}&{x^2} + {y^2} + {z^2} = \left( {\begin{array}{*{20}{c}}x&y&z\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{t_1}}&{{t_2}}&{{t_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{b{y_1}}&{c{z_1}}\\{a{x_2}}&{b{y_2}}&{c{z_2}}\\{a{x_3}}&{b{y_3}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{a{x_1}}&{a{x_2}}&{a{x_3}}\\{b{y_1}}&{b{y_2}}&{b{y_3}}\\{c{z_1}}&{c{z_2}}&{c{z_3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{t_1}}\\{{t_2}}\\{{t_3}}\end{array}} \right)\\=& \sum\limits_{i = 1}^3 {{t_i}^2\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)} = \sum\limits_{i = 1}^3 {\frac{1}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} = \sum\limits_{i = 1}^3 {\frac{{a{x_i}^2 + b{y_i}^2 + c{z_i}^2}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}} \\=& a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right)\\+& c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right).\end{align*}

注意到对任意的$w$,记$w = \sum\limits_{i = 1}^3 {{s_i}{n_i}} $,则${s_i} = \frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}$,即\[w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} .\]

分别令$w = \sum\limits_{i = 1}^3 {\frac{{w \cdot {n_i}}}{{\left( {{a^2}{x_i}^2 + {b^2}{y_i}^2 + {c^2}{z_i}^2} \right)}}{n_i}} $,得

\begin{align*}a\left( {\frac{{{x_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{x_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{x_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{a},\\b\left( {\frac{{{y_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{y_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{y_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{b},\\c\left( {\frac{{{z_1}^2}}{{\left( {{a^2}{x_1}^2 + {b^2}{y_1}^2 + {c^2}{z_1}^2} \right)}} + \frac{{{z_2}^2}}{{\left( {{a^2}{x_2}^2 + {b^2}{y_2}^2 + {c^2}{z_2}^2} \right)}} + \frac{{{z_3}^2}}{{\left( {{a^2}{x_3}^2 + {b^2}{y_3}^2 + {c^2}{z_3}^2} \right)}}} \right) &= \frac{1}{c}.\end{align*}

因此有\[{x^2} + {y^2} + {z^2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.\]

解法二.记$d_i=\sqrt{a^2x_i^2+b^2y_i^2+c^2z_i^2},\ i=1,2,3$,则$\begin{pmatrix}\frac{ax_1}{d_1}&\frac{by_1}{d_1}&\frac{cz_1}{d_1}\\\frac{ax_2}{d_2}&\frac{by_2}{d_2}&\frac{cz_2}{d_2}\\\frac{ax_3}{d_3}&\frac{by_3}{d_3}&\frac{cz_3}{d_3}\end{pmatrix} $是正交矩阵,从而有

\[\frac{a^2x_1^2}{d_1^2}+\frac{a^2x_2^2}{d_2^2}+\frac{a^3x_3^2}{d_3^2}=1\Rightarrow \frac{ax_1^2}{d_1^2}+\frac{ax_2^2}{d_2^2}+\frac{ax_3^2}{d_3^2}=\frac{1}{a}.\]

类似得到另外两式, 相加便有

\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ax_1^2+by_1^2+cz_1^2}{d_1^2}+\frac{ax_2^2+by_2^2+cz_2^2}{d_2^2}+\frac{ax_3^2+by_3^2+cz_3^2}{d_3^2}=\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}.\]

网友WB问我一道积分题\[\int_0^\infty  {\int_0^\infty  {{e^{ - a\sqrt {{x^2} + {y^2}} }}\cos \alpha x\cos \beta ydxdy} } .\]

经极坐标代换后得到

\[\int_0^\infty  {dr} \int_0^{\frac{\pi }{2}} {r{e^{ - ar}}\cos \left( {\alpha r\cos \theta } \right)\cos \left( {\beta r\sin \theta } \right)d\theta } .\]

然后利用积化和差公式感觉可以继续往下算.

计算$$\displaystyle\lim_{n\rightarrow\infty}\left({2\sqrt n}-\sum_{k=1}^n\frac{1}{\sqrt k}\right).$$

证明\[\sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor  + 1} \right)\pi }}{{2n}}} \right)}  = \cot \left( {\frac{\pi }{{2n}}} \right)\cos \left( {\frac{\pi }{{2n}}} \right),\]

where $\lfloor\cdot\rfloor$ is greatest integer function.

计算$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx}.$$

第一个积分的解答:

http://math.stackexchange.com/questions/958624/prove-that-int-01-sin-pi-xxx1-x1-x-dx-frac-pi-e24

设$A\in M_n (\mathbb{R})$对称正定, $X\in M_{n\times m} (\mathbb{R})$且$X^TX=I_m$.证明$X^TA^{-1}X-{X^TAX}^{-1}$是半正定矩阵.

证.首先

\[\left( {\begin{array}{*{20}{c}}{{X^T}AX}&{{I_m}}\\{{I_m}}&{{X^T}{A^{ - 1}}X}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{X^T}AX}&{{X^T}X}\\{{X^T}X}&{{X^T}{A^{ - 1}}X}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{X^T}}&0\\0&{{X^T}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}A&{{I_n}}\\{{I_n}}&{{A^{ - 1}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}X&0\\0&X\end{array}} \right).\]

而

\[\left( {\begin{array}{*{20}{c}}{{I_n}}&0\\{ - {A^{ - 1}}}&{{I_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}A&{{I_n}}\\{{I_n}}&{{A^{ - 1}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{I_n}}&{ - {A^{ - 1}}}\\0&{{I_n}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}A&0\\0&0\end{array}} \right).\]

所以$\left( {\begin{array}{*{20}{c}}A&{{I_n}}\\{{I_n}}&{{A^{ - 1}}}\end{array}} \right)$半正定,所以$\left( {\begin{array}{*{20}{c}}{{X^T}AX}&{{I_m}}\\{{I_m}}&{{X^T}{A^{ - 1}}X}\end{array}} \right)$半正定.

又\[\left( {\begin{array}{*{20}{c}}{{I_m}}&0\\{ - {{\left( {{X^T}AX} \right)}^{ - 1}}}&{{I_m}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{X^T}AX}&{{I_m}}\\{{I_m}}&{{X^T}{A^{ - 1}}X}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{I_m}}&{ - {{\left( {{X^T}AX} \right)}^{ - 1}}}\\0&{{I_m}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{X^T}AX}&0\\0&{{X^T}{A^{ - 1}}X - {{\left( {{X^T}AX} \right)}^{ - 1}}}\end{array}} \right),\]

所以$X^TA^{-1}X-{X^TAX}^{-1}$是半正定矩阵.

设 $x_i>0,i=1,2,\cdots$, 证明矩阵 $${\left( {\frac{{\ln \left( {1 + {x_i} + {x_j}} \right) - \ln \left( {1 + \left| {{x_i} - {x_j}} \right|} \right)}}{{{x_i} + {x_j} - \left| {{x_i} - {x_j}} \right|}}} \right)_{n \times n}} $$是半正定矩阵.

证.(morrismodel)这是一道合成题, 我把它分解开来:

题1. 当$0<x_1<x_2<\cdots<x_n$时,

$$\left(\min\{x_i,x_j\}\right)_{n\times n}= \begin{pmatrix} x_1&x_1&x_1&\cdots&x_1\\ x_1&x_2&x_2&\cdots &x_2\\ x_1&x_2&x_3&\cdots&x_3\\ \cdots&\cdots&\cdots&\cdots\\ x_1&x_2&x_3&\cdots&x_n \end{pmatrix} $$

正定.

证明. 由如下的矩阵恒等式得到:

\begin{align*}&\begin{pmatrix} a_1&\\ a_1&a_2\\ a_1&a_2&a_3\\ \cdots&\cdots&\cdots&\cdots\\ a_1&a_2&a_3&\cdots&a_n \end{pmatrix} \begin{pmatrix} a_1&a_1&a_1&\cdots&a_1\\ 0&a_2&a_2&\cdots &a_2\\ 0&0&a_3&\cdots&a_3\\ \cdots&\cdots&\cdots&\cdots\\ 0&0&0&\cdots&a_n \end{pmatrix}\\=& \begin{pmatrix} a_1^2&a_1^2&a_1^2&\cdots&a_1^2\\ a_1^2&a_1^2+a_2^2&a_1^2+a_2^2&\cdots &a_1^2+a_2^2\\ a_1^2&a_1^2+a_2^2&a_1^2+a_2^2+a_3^2&\cdots&a_1^2+a_2^2+a_3^2\\ \cdots&\cdots&\cdots&\cdots\\ a_1^2&a_1^2+a_2^2&a_1^2+a_2^2+a_3^2&\cdots&a_1^2+a_2^2+a_3^2+\cdots+a_n^2 \end{pmatrix}.\end{align*}

题2. 当正数$x_1,x_2,\cdots,x_n$两两不相等时,

$$\left(\min\{x_i,x_j\}\right)_{n\times n}= \begin{pmatrix} x_1&x_1&x_1&\cdots&x_1\\ x_1&x_2&x_2&\cdots &x_2\\ x_1&x_2&x_3&\cdots&x_3\\ \cdots&\cdots&\cdots&\cdots\\ x_1&x_2&x_3&\cdots&x_n \end{pmatrix} $$

正定.

证明. 对任何$N$阶置换$\sigma\in S_N$,

\begin{align*} &\sum_{i,j=1}^Na_{i}a_{j}\min\{x_{\sigma(i)},x_{\sigma(j)}\}=\sum_{i=1}^N\left(\sum_{j=1}^Na_{i}a_{j}\min\{x_{\sigma(i)},x_{\sigma(j)}\}\right)\\=&\sum_{i=1}^N\left(\sum_{j=1}^Na_{i}a_{\sigma^{-1}(j)}\min\{x_{\sigma(i)},x_{j}\}\right)=\sum_{j=1}^N\left(\sum_{i=1}^Na_{i}a_{\sigma^{-1}(j)}\min\{x_{\sigma(i)},x_{j}\}\right)\\=&\sum_{j=1}^N\left(\sum_{i=1}^Na_{\sigma^{-1}(i)}a_{\sigma^{-1}(j)}\min\{x_{i},x_{j}\}\right)=\sum_{i,j=1}^Na_{\sigma^{-1}(i)}a_{\sigma^{-1}(j)}\min\{x_{i},x_{j}\}. \end{align*}

再结合题1就得到结论.

题3. 当正数$x_1,x_2,\cdots,x_n$两两不相等时,

$$\left(e^{x_i+x_j-|x_i-x_j|}\right)_{n\times n}= \begin{pmatrix} e^{2x_1}&e^{2x_1}&e^{2x_1}&\cdots&e^{2x_1}\\ e^{2x_1}&e^{2x_2}&e^{2x_2}&\cdots &e^{2x_2}\\ e^{2x_1}&e^{2x_2}&e^{2x_3}&\cdots&e^{2x_3}\\ \cdots&\cdots&\cdots&\cdots\\ e^{2x_1}&e^{2x_2}&e^{2x_3}&\cdots&e^{2x_n} \end{pmatrix} $$

正定.

证明. 由题2得到.

题4. 当正数$x_1,x_2,\cdots,x_n$两两不相等且$\theta\in[0,1]$时,

$$\left(\frac{1}{1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|}\right)_{n\times n} $$

正定.

证明.

\begin{align*} &\sum_{i,j=1}^N\frac{a_ia_j}{1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|}\\=&\sum_{i,j=1}^Na_ia_j\int_0^\infty e^{-t(1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|)}dt\\ =&\int_0^\infty e^{-t}\left(\sum_{i,j=1}^N(a_ie^{-tx_i})(a_je^{-tx_j})e^{t(1-\theta)(x_i+x_j-|x_i-x_j|)}\right)dt. \end{align*}

再由题3得到结论.

题5. 当正数$x_1,x_2,\cdots,x_n$两两不相等时,

$$\left( {\frac{{\ln \left( {1 + {x_i} + {x_j}} \right) - \ln \left( {1 + \left| {{x_i} - {x_j}} \right|} \right)}}{{{x_i} + {x_j} - \left| {{x_i} - {x_j}} \right|}}} \right)_{n \times n} $$

正定.

证明.

$${\frac{{\ln \left( {1 + {x_i} + {x_j}} \right) - \ln \left( {1 + \left| {{x_i} - {x_j}} \right|} \right)}}{{{x_i} + {x_j} - \left| {{x_i} - {x_j}}\right|}}}=\int_0^1 \frac{1}{1+\theta(x_i+x_j)+(1-\theta)|x_i-x_j|} d\theta.$$

再由题4得到结论.

回到原题, 没有假设正数$x_i$两两不等, 只需在题5的基础上摄动一下就能得到半正定性.

博士论坛群里ZHX问的一道平面几何题，由余神提示，ZHX自己整理的解答：