一道很有趣的多元函数求极值问题）
本文部分地方存在计算错误，但思路是正确的，后面等我有时间改一下
题目
设 0 < a < b , n ⩾ 1 0<a<b,n\geqslant 1 0<a<b,n⩾1. 试在 ( a , b ) (a,b) (a,b) 中选取 n n n 个点 x 1 , x 2 , ⋯ , x n x_1,x_2,\cdots,x_n x1,x2,⋯,xn 使得：
f = x 1 x 2 ⋯ x n ( a + x 1 ) ( x 1 + x 2 ) ⋯ ( x n + b ) (1) f=\frac{x_1x_2\cdots x_n}{(a+x_1)(x_1+x_2)\cdots(x_n+b)}\tag{1} f=(a+x1)(x1+x2)⋯(xn+b)x1x2⋯xn(1)
取最大值
一、求雅克比（Jacobian）矩阵
为了书写简便，我们令
f 1 ≜ x 1 x 2 ⋯ x n f 2 ≜ ( a + x 1 ) ( x 1 + x 2 ) + ⋯ + ( x n + b ) \begin{aligned} f_1 &\triangleq x_1x_2\cdots x_n\\ f_2&\triangleq(a+x_1)(x_1+x_2)+\cdots+(x_n+b) \end{aligned} f1f2≜x1x2⋯xn≜(a+x1)(x1+x2)+⋯+(xn+b)
考虑
∂ f ∂ x i i = 1 , 2 , ⋯ n \frac{\partial f}{\partial x_i}\qquad i=1,2,\cdots n ∂xi∂fi=1,2,⋯n
当 i = 1 i =1 i=1 时：
∂ f ∂ x 1 = ∂ f 1 x 1 f 2 − f 1 ∂ f 2 x 1 f 2 2 = f 1 x 1 f 2 − f 1 ( a + 2 x 1 + x 2 ) f 2 ( a + x 1 ) ( x 1 + x 2 ) f 2 2 = f 1 x 1 f 2 − f 1 f 2 a + 2 x 1 + x 2 ( a + x 1 ) ( x 1 + x 2 ) = f x 1 − f ( a + x 1 ( a + x 1 ) ( x 1 + x 2 ) + x 1 + x 2 ( a + x 1 ) ( x 1 + x 2 ) ) = f ( 1 x 1 − 1 x 1 + x 2 − 1 a + x 1 ) \begin{aligned} \dfrac{\partial f}{\partial x_1}&=\dfrac{\dfrac{\partial f_1}{x_1}}{f_2}-f_1\dfrac{\dfrac{\partial f_2}{x_1}}{f_2^2}\\ &=\dfrac{\dfrac{f_1}{x_1}}{f_2}-f_1\dfrac{(a+2x_1+x_2)\dfrac{f_2}{(a+x_1)(x_1+x_2)}}{f_2^2}\\ &=\dfrac{f_1}{x_1f_2}-\dfrac{f_1}{f_2}\frac{a+2x_1+x_2}{(a+x_1)(x_1+x_2)}\\ &=\frac{f}{x_1}-f\left(\frac{a+x_1}{(a+x_1)(x_1+x_2)}+\frac{x_1+x_2}{(a+x_1)(x_1+x_2)}\right)\\ &=f(\frac{1}{x_1}-\frac{1}{x_1+x_2}-\frac{1}{a+x_1}) \end{aligned} ∂x1∂f=f2x1∂f1−f1f22x1∂f2=f2x1f1−f1f22(a+2x1+x2)(a+x1)(x1+x2)f2=x1f2f1−f2f1(a+x1)(x1+x2)a+2x1+x2=x1f−f((a+x1)(x1+x2)a+x1+(a+x1)(x1+x2)x1+x2)=f(x11−x1+x21−a+x11)
当 2 ⩽ i ⩽ n − 1 2\leqslant i\leqslant n-1 2⩽i⩽n−1 时：
∂ f ∂ x i = ∂ f 1 x i f 2 − f 1 ∂ f 2 x i f 2 2 = f 1 x i f 2 − f 1 ( x i − 1 + 2 x i + x i + 1 ) f 2 ( x i − 1 + x i ) ( x i + x i + 1 ) f 2 2 = f 1 x i f 2 − f 1 f 2 x i − 1 + 2 x i + x i + 1 ( x i − 1 + x i ) ( x i + x i + 1 ) = f x i − f ( x i − 1 + x i ( x i − 1 + x i ) ( x i + x i + 1 ) + x i + x i + 1 ( x i − 1 + x i ) ( x i + x i + 1 ) ) = f ( 1 x i − 1 x i − 1 + x i − 1 x i + x i + 1 ) \begin{aligned} \dfrac{\partial f}{\partial x_i}&=\dfrac{\dfrac{\partial f_1}{x_i}}{f_2}-f_1\dfrac{\dfrac{\partial f_2}{x_i}}{f_2^2}\\ &=\dfrac{\dfrac{f_1}{x_i}}{f_2}-f_1\dfrac{(x_{i-1}+2x_i+x_{i+1})\dfrac{f_2}{(x_{i-1}+x_{i})(x_i+x_{i+1})}}{f_2^2}\\ &=\dfrac{f_1}{x_if_2}-\dfrac{f_1}{f_2}\frac{x_{i-1}+2x_i+x_{i+1}}{(x_{i-1}+x_i)(x_i+x_{i+1})}\\ &=\frac{f}{x_i}-f\left(\frac{x_{i-1}+x_i}{(x_{i-1}+x_i)(x_i+x_{i+1})}+\frac{x_i+x_{i+1}}{(x_{i-1}+x_i)(x_i+x_{i+1})}\right)\\ &=f(\frac{1}{x_i}-\frac{1}{x_{i-1}+x_i}-\frac{1}{x_i+x_{i+1}}) \end{aligned} ∂xi∂f=f2xi∂f1−f1f22xi∂f2=f2xif1−f1f22(xi−1+2xi+xi+1)(xi−1+xi)(xi+xi+1)f2=xif2f1−f2f1(xi−1+xi)(xi+xi+1)xi−1+2xi+xi+1=xif−f((xi−1+xi)(xi+xi+1)xi−1+xi+(xi−1+xi)(xi+xi+1)xi+xi+1)=f(xi1−xi−1+xi1−xi+xi+11)
当 i = n i=n i=n 时：
∂ f ∂ x n = ∂ f 1 x n f 2 − f 1 ∂ f 2 x n f 2 2 = f 1 x n f 2 − f 1 ( x n − 1 + 2 x n + b ) f 2 ( x n − 1 + x n ) ( x n + b ) f 2 2 = f 1 x n f 2 − f 1 f 2 x n − 1 + 2 x n + b ( x n − 1 + x n ) ( x n + b ) = f x n − f ( x n − 1 + x n ( x n − 1 + x n ) ( x n + x n + 1 ) + x n + b ( x n − 1 + x n ) ( x n + b ) ) = f ( 1 x n − 1 x n − 1 + x n − 1 x n + b ) \begin{aligned} \dfrac{\partial f}{\partial x_n}&=\dfrac{\dfrac{\partial f_1}{x_n}}{f_2}-f_1\dfrac{\dfrac{\partial f_2}{x_n}}{f_2^2}\\ &=\dfrac{\dfrac{f_1}{x_n}}{f_2}-f_1\dfrac{(x_{n-1}+2x_n+b)\dfrac{f_2}{(x_{n-1}+x_{n})(x_n+b)}}{f_2^2}\\ &=\dfrac{f_1}{x_nf_2}-\dfrac{f_1}{f_2}\frac{x_{n-1}+2x_n+b}{(x_{n-1}+x_n)(x_n+b)}\\ &=\frac{f}{x_n}-f\left(\frac{x_{n-1}+x_n}{(x_{n-1}+x_n)(x_n+x_{n+1})}+\frac{x_n+b}{(x_{n-1}+x_n)(x_n+b)}\right)\\ &=f(\frac{1}{x_n}-\frac{1}{x_{n-1}+x_n}-\frac{1}{x_n+b}) \end{aligned} ∂xn∂f=f2xn∂f1−f1f22xn∂f2=f2xnf1−f1f22(xn−1+2xn+b)(xn−1+xn)(xn+b)f2=xnf2f1−f2f1(xn−1+xn)(xn+b)xn−1+2xn+b=xnf−f((xn−1+xn)(xn+xn+1)xn−1+xn+(xn−1+xn)(xn+b)xn+b)=f(xn1−xn−1+xn1−xn+b1)
所以，
D f = f [ − 1 a + x 1 + 1 x 1 − 1 x 1 + x 2 − 1 x 1 + x 2 + 1 x 2 − 1 x 2 + x 3 ⋮ − 1 x n − 1 + x n + 1 x n − 1 x n + b ] {\rm D}f=f\begin{bmatrix} -\frac{1}{a+x_1}+\frac{1}{x_1}-\frac{1}{x_1+x_2}\\[5pt] -\frac{1}{x_{1}+x_2}+\frac{1}{x_2}-\frac{1}{x_2+x_{3}}\\[5pt] \vdots\\[5pt] -\frac{1}{x_{n-1}+x_n}+\frac{1}{x_n}-\frac{1}{x_n+b} \end{bmatrix} Df=f⎣⎢⎢⎢⎢⎢⎢⎡−a+x11+x11−x1+x21−x1+x21+x21−x2+x31⋮−xn−1+xn1+xn1−xn+b1⎦⎥⎥⎥⎥⎥⎥⎤
二、求驻点
令 D f = 0 {\rm D}f=0 Df=0
D f = f [ − 1 a + x 1 + 1 x 1 − 1 x 1 + x 2 − 1 x 1 + x 2 + 1 x 2 − 1 x 2 + x 3 ⋮ − 1 x n − 1 + x n + 1 x n − 1 x n + b ] = 0 {\rm D}f=f\begin{bmatrix} -\frac{1}{a+x_1}+\frac{1}{x_1}-\frac{1}{x_1+x_2}\\[5pt] -\frac{1}{x_{1}+x_2}+\frac{1}{x_2}-\frac{1}{x_2+x_{3}}\\[5pt] \vdots\\[5pt] -\frac{1}{x_{n-1}+x_n}+\frac{1}{x_n}-\frac{1}{x_n+b} \end{bmatrix}=0 Df=f⎣⎢⎢⎢⎢⎢⎢⎡−a+x11+x11−x1+x21−x1+x21+x21−x2+x31⋮−xn−1+xn1+xn1−xn+b1⎦⎥⎥⎥⎥⎥⎥⎤=0
得到：
− 1 a + x 1 + 1 x 1 − 1 x 1 + x 2 = 0 − 1 x 1 + x 2 + 1 x 2 − 1 x 2 + x 3 = 0 − 1 x 2 + x 3 + 1 x 3 − 1 x 3 + x 4 = 0 ⋮ − 1 x n − 1 + x n + 1 x n − 1 x n + b = 0 -\frac{1}{a+x_1}+\frac{1}{x_1}-\frac{1}{x_1+x_2}=0\\[5pt] -\frac{1}{x_{1}+x_2}+\frac{1}{x_2}-\frac{1}{x_2+x_{3}}=0\\[5pt] -\frac{1}{x_{2}+x_3}+\frac{1}{x_3}-\frac{1}{x_3+x_{4}}=0\\[5pt] \vdots\\[5pt] -\frac{1}{x_{n-1}+x_n}+\frac{1}{x_n}-\frac{1}{x_n+b}=0 −a+x11+x11−x1+x21=0−x1+x21+x21−x2+x31=0−x2+x31+x31−x3+x41=0⋮−xn−1+xn1+xn1−xn+b1=0
整理得到：
x 2 x 1 = x 1 a x 3 x 2 = x 2 x 1 ⋮ ⋮ x n x n − 1 = x n − 1 x i − 2 b x n = x n x n − 1 \frac{x_2}{x_1}=\frac{x_1}{a}\\[5pt] \frac{x_3}{x_2}=\frac{x_2}{x_1}\\[5pt] \vdots\\[5pt] \vdots\\[5pt] \frac{x_n}{x_{n-1}}=\frac{x_{n-1}}{x_{i-2}}\\[5pt] \frac{b}{x_n}=\frac{x_n}{x_{n-1}} x1x2=ax1x2x3=x1x2⋮⋮xn−1xn=xi−2xn−1xnb=xn−1xn
不难看出， { x n } \{x_n\} { xn} 是一个等比数列
x n = k n a , k = a − 1 n + 1 b 1 n + 1 x_n=k^na,\qquad k=a^{-\frac{1}{n+1}}b^\frac{1}{n+1} xn=kna,k=a−n+11bn+11
记
X ≜ ( x 1 , x 2 , ⋯ , x n ) \mathbb{X}\triangleq(x_1,x_2,\cdots,x_n) X≜(x1,x2,⋯,xn)
三、计算驻点处函数值
f ∣ X = k a ⋅ k 2 a ⋯ k n a ( a + k a ) ( k a + k 2 a ) ⋯ ( k n a + k n + 1 a ) = k 1 + 2 + ⋯ + n a n k 1 + 2 + ⋯ + n a n + 1 ( 1 + k ) n + 1 = 1 a ( 1 + k ) n + 1 \begin{aligned} f|_\mathbb{X}&=\dfrac{ka\cdot k^2a\cdots k^na}{(a+ka)(ka+k^2a)\cdots(k^na+k^{n+1}a)}\\[10pt] &=\frac{k^{1+2+\cdots+n}a^n}{k^{1+2+\cdots+n}a^{n+1}(1+k)^{n+1}}\\[10pt] &=\frac{1}{a(1+k)^{n+1}} \end{aligned} f∣X=(a+ka)(ka+k2a)⋯(kna+kn+1a)ka⋅k2a⋯kna=k1+2+⋯+nan+1(1+k)n+1k1+2+⋯+nan=a(1+k)n+11
四、计算驻点处黑塞（Hessian）矩阵
计算
∂ 2 f ∂ x i ∂ x j \frac{\partial^2f}{\partial x_i\partial x_j} ∂xi∂xj∂2f
因为是对称矩阵，所以无需考虑 i > j i>j i>j 的情况
当 i = 1 i=1 i=1 时：
\qquad 当 j = 1 j=1 j=1 时：
∂ 2 f ∂ x 1 2 = ∂ ∂ x 1 ∂ f ∂ x 1 = ∂ ∂ x 1 [ f ( 1 x 1 − 1 x 1 + x 2 − 1 a + x 1 ) ] = f 1 ( 1 x 1 − 1 x 1 + x 2 − 1 a + x 1 ) + f ( − 1 x 1 2 + 1 ( x 1 + x 2 ) 2 + 1 ( a + x 1 ) 2 \begin{aligned} \frac{\partial^2f}{\partial x_1^2}&=\frac{\partial}{\partial x_1}\frac{\partial f}{\partial x_1}\\ &=\frac{\partial}{\partial x_1}[f(\frac{1}{x_1}-\frac{1}{x_1+x_2}-\frac{1}{a+x_1})]\\ &=f_1(\frac{1}{x_1}-\frac{1}{x_1+x_2}-\frac{1}{a+x_1})+f(-\frac{1}{x_1^2}+\frac{1}{(x_1+x_2)^2}+\frac{1}{(a+x_1)^2} \end{aligned} ∂x12∂2f=∂x1∂∂x1∂f=∂x1∂[f(x11−x1+x21−a+x11)]=f1(x11−x1+x21−a+x11)+f(−x121+(x1+x2)21+(a+x1)21
f 1 ( 1 x 1 − 1 x 1 + x 2 − 1 a + x 1 ) ∣ X = 0 f_1(\frac{1}{x_1}-\frac{1}{x_1+x_2}-\frac{1}{a+x_1})|_{\mathbb{X}}=0 f1(x11−x1+x21−a+x11)∣X=0
f ( − 1 x 1 2 + 1 ( x 1 + x 2 ) 2 + 1 ( a + x 1 ) 2 ) ∣ X = 1 a 1 ( 1 + k ) n + 1 ( 1 x 1 2 ) ( − 1 + 1 ( 1 + x 2 x 1 ) 2 + 1 ( 1 + a x 1 ) 2 ) = 1 ( 1 + k ) n + 1 1 a x 1 2 ( x 1 2 ( x 1 + a ) 2 + a 2 ( x 1 + a ) 2 − 1 ) = − 1 ( 1 + k ) n + 1 2 x 1 ( x 1 + a ) 2 \begin{aligned} f(-\frac{1}{x_1^2}+\frac{1}{(x_1+x_2)^2}+\frac{1}{(a+x_1)^2})|_{\mathbb{X}}&=\frac{1}{a}\frac{1}{(1+k)^{n+1}}(\frac{1}{x_1^2})(-1+\frac{1}{(1+\frac{x_2}{x_1})^2}+\frac{1}{(1+\frac{a}{x_1})^2})\\[10pt] &=\frac{1}{(1+k)^{n+1}}\frac{1}{ax_1^2}(\frac{x_1^2}{(x_1+a)^2}+\frac{a^2}{(x_1+a)^2}-1)\\[10pt] &=-\frac{1}{(1+k)^{n+1}}\frac{2}{x_1(x_1+a)^2} \end{aligned} f(−x121+(x1+x2)21+(a+x1)21)∣X=a1(1+k)n+11(x121)(−1+(1+x1x2)21+(1+x1a)21)=(1+k)n+11ax121((x1+a)2x12+(x1+a)2a2−1)=−(1+k)n+11x1(x1+a)22
\qquad 当 j = 2 j=2 j=2 时：
∂ 2 f ∂ x 1 ∂ x 2 = ∂ ∂ x 2 ∂ f ∂ x 1 = ∂ ∂ x 2 [ f ( 1 x 1 − 1 x 1 + x 2 − 1 a + x 1 ) ] = f 2 ( 1 x 1 − 1 x 1 + x 2 − 1 a + x 1 ) + f 1 ( x 1 + x 2 ) 2 \begin{aligned} \frac{\partial^2f}{\partial x_1\partial x_2}&=\frac{\partial}{\partial x_2}\frac{\partial f}{\partial x_1}\\ &=\frac{\partial}{\partial x_2}[f(\frac{1}{x_1}-\frac{1}{x_1+x_2}-\frac{1}{a+x_1})]\\ &=f_2(\frac{1}{x_1}-\frac{1}{x_1+x_2}-\frac{1}{a+x_1})+f\frac{1}{(x_1+x_2)^2} \end{aligned} ∂x1∂x2∂2f=∂x2∂∂x1∂f=∂x2∂[f(x11−x1+x21−a+x11)]=f2(x11−x1+x21−a+x11)+f(x1+x2)21
∂ 2 f ∂ x 1 ∂ x 2 ∣ X = 0 + 1 a 1 ( 1 + k ) n + 1 1 ( x 1 + x 2 ) 2 = 1 a ( x 1 + x 2 ) 2 1 ( 1 + k ) n + 1 \begin{aligned} \frac{\partial^2f}{\partial x_1\partial x_2}|_\mathbb{X}&=0+\frac{1}{a}\frac{1}{(1+k)^{n+1}}\frac{1}{(x_1+x_2)^2}\\[10pt] &=\frac{1}{a(x_1+x_2)^2}\frac{1}{(1+k)^{n+1}} \end{aligned} ∂x1∂x2∂2f∣X=0+a1(1+k)n+11(x1+x2)21=a(x1+x2)21(1+k)n+11
\qquad 当 j = 3 , 4 , ⋯ , n j=3,4,\cdots,n j=3,4,⋯,n
∂ 2 f ∂ x i ∂ x j = 0 \frac{\partial^2f}{\partial x_i\partial x_j}=0 ∂xi∂xj∂2f=0
当 i = 2 , 3 , ⋯ , n − 1 i=2,3,\cdots,n-1 i=2,3,⋯,n−1
\qquad 当 j = i j=i j=i
∂ 2 f ∂ x i 2 = ∂ ∂ x i ∂ f ∂ x i = ∂ ∂ x i [ f ( 1 x i − 1 x i − 1 + x i − 1 x i + x i + 1 ) ] = f i ( 1 x i − 1 x i − 1 + x i − 1 x i + x i + 1 ) + f ( − 1 x i 2 + 1 ( x i − 1 + x i ) 2 + 1 ( x i + x i + 1 ) 2 ) \begin{aligned} \frac{\partial^2f}{\partial x_i^2}&=\frac{\partial}{\partial x_i}\frac{\partial f}{\partial x_i}\\ &=\frac{\partial}{\partial x_i}[f(\frac{1}{x_i}-\frac{1}{x_{i-1}+x_i}-\frac{1}{x_i+x_{i+1}})]\\ &=f_i(\frac{1}{x_i}-\frac{1}{x_{i-1}+x_i}-\frac{1}{x_i+x_{i+1}})+f(-\frac{1}{x_i^2}+\frac{1}{(x_{i-1}+x_i)^2}+\frac{1}{(x_i+x_{i+1})^2}) \end{aligned} ∂xi2∂2f=∂xi∂∂xi∂f=∂xi∂[f(xi1−xi−1+xi1−xi+xi+11)]=fi(xi1−xi−1+xi1−xi+xi+11)+f(−xi21+(xi−1+xi)21+(xi+xi+1)21)
f i ( 1 x i − 1 x i − 1 + x i − 1 x i + x i + 1 ) ∣ X = 0 f_i(\frac{1}{x_i}-\frac{1}{x_{i-1}+x_i}-\frac{1}{x_i+x_{i+1}})|_\mathbb{X}=0 fi(xi1−xi−1+xi1−xi+xi+11)∣X=0
f ( − 1 x i 2 + 1 ( x i − 1 + x i ) 2 + 1 ( x i + x i + 1 ) 2 ) ∣ X = 1 a 1 ( 1 + k ) n + 1 1 x i 2 ( − 1 + 1 ( 1 + x i − 1 x i ) 2 + 1 ( 1 + x i + 1 x i ) 2 ) = 1 a x i 2 1 ( 1 + k ) n + 1 − 2 a x 1 ( a + x 1 ) 2 = − 2 x 1 x i 2 ( a + x 1 ) 2 1 ( 1 + k ) n + 1 \begin{aligned} f(-\frac{1}{x_i^2}+\frac{1}{(x_{i-1}+x_i)^2}+\frac{1}{(x_i+x_{i+1})^2})|_\mathbb{X}&=\frac{1}{a}\frac{1}{(1+k)^{n+1}}\frac{1}{x_i^2}(-1+\frac{1}{(1+\frac{x_{i-1}}{x_i})^2}+\frac{1}{(1+\frac{x_{i+1}}{x_{i}})^2})\\[10pt] &=\frac{1}{ax_i^2}\frac{1}{(1+k)^{n+1}}\frac{-2ax_1}{(a+x_1)^2}\\[10pt] &=\frac{-2x_1}{x_i^2(a+x_1)^2}\frac{1}{(1+k)^{n+1}} \end{aligned} f(−xi21+(xi−1+xi)21+(xi+xi+1)21)∣X=a1(1+k)n+11xi21(−1+(1+xixi−1)21+(1+xixi+1)21)=axi21(1+k)n+11(a+x1)2−2ax1=xi2(a+x1)2−2x1(1+k)n+11
\qquad 当 j = i + 1 j=i+1 j=i+1 时
∂ 2 f ∂ x i + 1 ∂ x i = ∂ ∂ x i + 1 ∂ f ∂ x i = ∂ ∂ x i + 1 [ f ( 1 x i − 1 x i − 1 + x i − 1 x i + x i + 1 ) ] = f i + 1 ( 1 x i − 1 x i − 1 + x i − 1 x i + x i + 1 ) + f 1 ( x i + x i + 1 ) 2 \begin{aligned} \frac{\partial^2f}{\partial x_{i+1}\partial x_i}&=\frac{\partial}{\partial x_{i+1}}\frac{\partial f}{\partial x_i}\\ &=\frac{\partial}{\partial x_{i+1}}[f(\frac{1}{x_i}-\frac{1}{x_{i-1}+x_i}-\frac{1}{x_i+x_{i+1}})]\\ &=f_{i+1}(\frac{1}{x_i}-\frac{1}{x_{i-1}+x_i}-\frac{1}{x_i+x_{i+1}})+f\frac{1}{(x_i+x_{i+1})^2} \end{aligned} ∂xi+1∂xi∂2f=∂xi+1∂∂xi∂f=∂xi+1∂[f(xi1−xi−1+xi1−xi+xi+11)]=fi+1(xi1−xi−1+xi1−xi+xi+11)+f(xi+xi+1)21
∂ 2 f ∂ x i + 1 ∂ x i ∣ X = 0 + 1 a 1 ( 1 + k ) n + 1 1 ( x i + x i + 1 ) 2 = 1 a ( x i + x i + 1 ) 2 1 ( 1 + k ) n + 1 \begin{aligned} \frac{\partial^2f}{\partial x_{i+_1}\partial x_i}|_\mathbb{X}&=0+\frac{1}{a}\frac{1}{(1+k)^{n+1}}\frac{1}{(x_i+x_{i+1})^2}\\[10pt] &=\frac{1}{a(x_i+x_{i+1})^2}\frac{1}{(1+k)^{n+1}} \end{aligned} ∂xi+1∂xi∂2f∣X=0+a1(1+k)n+11(xi+xi+1)21=a(xi+xi+1)21(1+k)n+11
\qquad 当 j = i + 2 , i + 3 , ⋯ , n j=i+2,i+3,\cdots,n j=i+2,i+3,⋯,n
∂ 2 f ∂ x i ∂ x j = 0 \frac{\partial^2f}{\partial x_i\partial x_j}=0 ∂xi∂xj∂2f=0
当 i = j = n i=j=n i=j=n 时：
∂ 2 f ∂ x n 2 = ∂ ∂ x n ∂ f ∂ x n = ∂ ∂ x n [ f ( 1 x n − 1 x n − 1 + x n − 1 x n + b ) ] = f n ( 1 x n − 1 x n − 1 + x n − 1 x n + b ) + f ( − 1 x n 2 + 1 ( x n − 1 + x n ) 2 + 1 ( x n + b ) 2 ) \begin{aligned} \frac{\partial^2f}{\partial x_n^2}&=\frac{\partial}{\partial x_n}\frac{\partial f}{\partial x_n}\\ &=\frac{\partial}{\partial x_n}[f(\frac{1}{x_n}-\frac{1}{x_{n-1}+x_n}-\frac{1}{x_n+b})]\\ &=f_n(\frac{1}{x_n}-\frac{1}{x_{n-1}+x_n}-\frac{1}{x_n+b})+f(-\frac{1}{x_n^2}+\frac{1}{(x_{n-1}+x_n)^2}+\frac{1}{(x_n+b)^2}) \end{aligned} ∂xn2∂2f=∂xn∂∂xn∂f=∂xn∂[f(xn1−xn−1+xn1−xn+b1)]=fn(xn1−xn−1+xn1−xn+b1)+f(−xn21+(xn−1+xn)21+(xn+b)21)
f n ( 1 x n − 1 x n − 1 + x i − 1 x n + b ) ∣ X = 0 f_n(\frac{1}{x_n}-\frac{1}{x_{n-1}+x_i}-\frac{1}{x_n+b})|_\mathbb{X}=0 fn(xn1−xn−1+xi1−xn+b1)∣X=0
f ( − 1 x n 2 + 1 ( x n − 1 + x n ) 2 + 1 ( x n + b ) 2 ) ∣ X = 1 a 1 ( 1 + k ) n + 1 1 x n 2 ( − 1 + 1 ( 1 + x n − 1 x n ) 2 + 1 ( 1 + b x n ) 2 ) = 1 a x n 2 − 2 a x 1 ( a + x 1 ) 2 1 ( 1 + k ) n + 1 = − 2 x 1 x n 2 ( a + x 1 ) 2 1 ( 1 + k ) n + 1 \begin{aligned} f(-\frac{1}{x_n^2}+\frac{1}{(x_{n-1}+x_n)^2}+\frac{1}{(x_n+b)^2})|_\mathbb{X}&=\frac{1}{a}\frac{1}{(1+k)^{n+1}}\frac{1}{x_n^2}(-1+\frac{1}{(1+\frac{x_{n-1}}{x_n})^2}+\frac{1}{(1+\frac{b}{x_{n}})^2})\\[10pt] &=\frac{1}{ax_n^2}\frac{-2ax_1}{(a+x_1)^2}\frac{1}{(1+k)^{n+1}}\\[10pt] &=\frac{-2x_1}{x_n^2(a+x_1)^2}\frac{1}{(1+k)^{n+1}} \end{aligned} f(−xn21+(xn−1+xn)21+(xn+b)21)∣X=a1(1+k)n+11xn21(−1+(1+xnxn−1)21+(1+xnb)21)=axn21(a+x1)2−2ax1(1+k)n+11=xn2(a+x1)2−2x1(1+k)n+11
综上，
H e s s f = 1 ( 1 + k ) n + 1 [ − 2 x 1 x 1 2 ( a + x 1 ) 2 1 a ( x 1 + x 2 ) 2 0 ⋯ 0 1 a ( x 1 + x 2 ) 2 − 2 x 1 x 2 2 ( a + x 1 ) 2 1 a ( x 2 + x 3 ) 2 ⋯ 0 0 1 a ( x 2 + x 3 ) 2 − 2 x 1 x 3 2 ( a + x 1 ) 2 ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ − 2 x 1 x n 2 ( a + x 1 ) 2 ] n × n {\rm Hess}f=\frac{1}{(1+k)^{n+1}}\begin{bmatrix} \frac{-2x_1}{x_1^2(a+x_1)^2} & \frac{1}{a(x_1+x_2)^2} & 0&\cdots &0\\[20pt] \frac{1}{a(x_1+x_2)^2} & \frac{-2x_1}{x_2^2(a+x_1)^2} &\frac{1}{a(x_2+x_{3})^2}&\cdots&0\\[20pt] 0&\frac{1}{a(x_2+x_{3})^2}& \frac{-2x_1}{x_3^2(a+x_1)^2} &\cdots&0\\[20pt] \vdots&\vdots&\vdots&&\vdots\\[20pt] 0&0&0&\cdots&\frac{-2x_1}{x_n^2(a+x_1)^2} \end{bmatrix}_{n\times n} Hessf=(1+k)n+11⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡x12(a+x1)2−2x1a(x1+x2)210⋮0a(x1+x2)21x22(a+x1)2−2x1a(x2+x3)21⋮00a(x2+x3)21x32(a+x1)2−2x1⋮0⋯⋯⋯⋯000⋮xn2(a+x1)2−2x1⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤n×n
五、证明驻点处黑塞（Hessian）矩阵为负定矩阵
考虑到 { x n } \{x_n\} { xn} 是等比数列，则 x n ≜ k n a x_n\triangleq k^na xn≜kna
H e s s {\rm Hess} Hess 矩阵可以改写成：
H e s s f = 1 ( 1 + k ) n + 1 [ − 2 a 3 k ( 1 + k ) 2 1 a 3 k 2 ( 1 + k ) 2 0 ⋯ 0 1 a 3 k 2 ( 1 + k ) 2 − 2 a 3 k 3 ( 1 + k ) 2 1 a 3 k 4 ( 1 + k ) 2 ⋯ 0 0 1 a 3 k 4 ( 1 + k ) 2 − 2 a 3 k 5 ( 1 + k ) 2 ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ − 2 a 3 k 2 n − 1 ( 1 + k ) 2 ] n × n {\rm Hess}f=\frac{1}{(1+k)^{n+1}}\begin{bmatrix} \frac{-2}{a^3k(1+k)^2} & \frac{1}{a^3k^2(1+k)^2} & 0&\cdots &0\\[20pt] \frac{1}{a^3k^2(1+k)^2} & \frac{-2}{a^3k^3(1+k)^2} &\frac{1}{a^3k^4(1+k)^2}&\cdots&0\\[20pt] 0&\frac{1}{a^3k^4(1+k)^2}& \frac{-2}{a^3k^5(1+k)^2} &\cdots&0\\[20pt] \vdots&\vdots&\vdots&&\vdots\\[20pt] 0&0&0&\cdots&\frac{-2}{a^3k^{2n-1}(1+k)^2} \end{bmatrix}_{n\times n} Hessf=(1+k)n+11⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡a3k(1+k)2−2a3k2(1+k)210⋮0a3k2(1+k)21a3k3(1+k)2−2a3k4(1+k)21⋮00a3k4(1+k)21a3k5(1+k)2−2⋮0⋯⋯⋯⋯000⋮a3k2n−1(1+k)2−2⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤n×n
因此，只需证明下列矩阵 A A A 是正定的
A = [ − 2 k 1 0 ⋯ 0 k 2 − 2 k 1 ⋯ 0 0 k 2 − 2 k ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ − 2 k ] n × n A=\begin{bmatrix} -2k & 1 & 0&\cdots &0\\[10pt] k^2 & -2k &1&\cdots&0\\[10pt] 0&k^2&-2k &\cdots&0\\[10pt] \vdots&\vdots&\vdots&&\vdots\\[10pt] 0&0&0&\cdots&-2k \end{bmatrix}_{n\times n} A=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡−2kk20⋮01−2kk2⋮001−2k⋮0⋯⋯⋯⋯000⋮−2k⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤n×n
设矩阵
D n = [ − 2 k 1 0 ⋯ 0 k 2 − 2 k 1 ⋯ 0 0 k 2 − 2 k ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ − 2 k ] n × n D_n=\begin{bmatrix} -2k & 1 & 0&\cdots &0\\[10pt] k^2 & -2k &1&\cdots&0\\[10pt] 0&k^2&-2k &\cdots&0\\[10pt] \vdots&\vdots&\vdots&&\vdots\\[10pt] 0&0&0&\cdots&-2k \end{bmatrix}_{n\times n} Dn=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡−2kk20⋮01−2kk2⋮001−2k⋮0⋯⋯⋯⋯000⋮−2k⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤n×n
假设 ∣ D n ∣ = ( − 1 ) n ( n + 1 ) k n |D_n|=(-1)^n(n+1)k^n ∣Dn∣=(−1)n(n+1)kn
当 n = 1 n=1 n=1 时,满足上式
假设 n ⩽ p n\leqslant p n⩽p 时,满足上式,则
D p = [ − 2 k 1 0 ⋯ 0 k 2 − 2 k 1 ⋯ 0 0 k 2 − 2 k ⋯ 0 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ − 2 k ] p × p = − 2 k D p − 1 + k 2 D p − 2 = ( − 1 ) p ( p + 1 ) k p \begin{aligned} D_p&=\begin{bmatrix} -2k & 1 & 0&\cdots &0\\[10pt] k^2 & -2k &1&\cdots&0\\[10pt] 0&k^2&-2k &\cdots&0\\[10pt] \vdots&\vdots&\vdots&&\vdots\\[10pt] 0&0&0&\cdots&-2k \end{bmatrix}_{p\times p}\\[10pt] &=-2kD_{p-1}+k^2D_{p-2}\\ &=(-1)^p(p+1)k^p \end{aligned} Dp=⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎡−2kk20⋮01−2kk2⋮001−2k⋮0⋯⋯⋯⋯000⋮−2k⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎤p×p=−2kDp−1+k2Dp−2=(−1)p(p+1)kp
满足上式
所以矩阵 A A A 的奇数主子式大于零，偶数主子式小于零，所以 A A A 是负定矩阵，所以 H e s s f {\rm Hess}f Hessf 是负定矩阵
结论
综上所述， f f f 在 ( k a , k 2 a , ⋯ , k n a ) (ka,k^2a,\cdots,k^na) (ka,k2a,⋯,kna) 处取得最小值 a / 2 a/2 a/2 ，其中 k = a − 1 n + 1 b 1 n + 1 k=a^{-\frac{1}{n+1}}b^\frac{1}{n+1} k=a−n+11bn+11