2017全国数学联合竞赛第10题

力工

各位看过来！画风变了耶！

kuing

很简单啊，首先通过消元后求导（又或者用拉格朗日乘数）算一下极值点，发现极值点有负数，可见最值只能在边界取得，于是对各变量分别取零看看如何，经过三回简单的二次函数分析比较后就能知道最小1最大9/5了。

心里有数之后，刚才这些分析全部扔掉，然后写过程，直接作差完事，即：

解：因为
\begin{gather*}
(x_1+3x_2+5x_3)\left(x_1+\frac{x_2}3+\frac{x_3}5\right)-(x_1+x_2+x_3)^2
=\frac4{15}(5x_1x_2+12x_1x_3+x_2x_3)\geqslant0,\\
(x_1+3x_2+5x_3)\left(x_1+\frac{x_2}3+\frac{x_3}5\right)-\frac95(x_1+x_2+x_3)^2
=-\frac4{15}\bigl(3(x_1-x_3)^2+x_2(x_1+3x_2+5x_3)\bigr)\leqslant0,
\end{gather*}
所以…………

v6mm131

回复 2# kuing

良心之答

游客

如果把其中一个固定，则目标是个开口向下的抛物线。

力工

大家好强！我觉得是不等式问题的风格有些变化，相对于前两年。

hbghlyj

$\iff$Minimize $x^TAx$ subject to $a^Tx=1$
where
$x=\pmatrix{x_1\\x_2\\x_3}$
$A=\pmatrix{1          & \frac53       & \frac{13}5    \\
\frac53    & 1             & \frac{17}{15} \\
\frac{13}5 & \frac{17}{15} & 1      }$
(union of two lines is a singular quadratic, $\operatorname{rank} A=2$, $A$ is singular)
$a=\pmatrix{1\\1\\1}$
cis page 457 (Definition 12.3.) quadratic constrained minimization problem
page 459
We shall prove that our constrained minimization problem has a unique solution given by the system of linear equations
$$\left(\begin{array}{cc}C^{-1} & A \\ A^{\top} & 0\end{array}\right)\left(\begin{array}{l}y \\ \lambda\end{array}\right)=\left(\begin{array}{l}b \\ f\end{array}\right)$$

hbghlyj

kuing 发表于 2017-9-10 10:07
很简单啊，首先通过消元后求导（又或者用拉格朗日乘数）算一下极值点，发现极值点有负数，可见最值只能在边 ...

objective function
$f(x) = x^TAx$
constraint function
$g(x) = a^Tx-1$
Lagrange multiplier
$L(x, \lambda) = f(x) - \lambda g(x)$
Calculate the derivative of the Lagrangian
$$\nabla L=2Ax-\lambda a$$
Solving $\nabla L=0$, we find:
$$x =\color{red}{\pmatrix{?\\?\\?}}$$
Substituting into $g(x)=0$, we get:
$$? = 0$$
Simplifying, we find:
$$\lambda = ?$$
Check the second-order conditions:
The Hessian matrix
$$H = \begin{bmatrix}
\frac{\partial^2 L}{\partial x_1^2} & \frac{\partial^2 L}{\partial x_1\partial x_2} & \frac{\partial^2 L}{\partial x_1\partial x_3} \\
\frac{\partial^2 L}{\partial x_2\partial x_1} & \frac{\partial^2 L}{\partial x_2^2} & \frac{\partial^2 L}{\partial x_2\partial x_3} \\
\frac{\partial^2 L}{\partial x_3\partial x_1} & \frac{\partial^2 L}{\partial x_3\partial x_2} & \frac{\partial^2 L}{\partial x_3^2}
\end{bmatrix}= ?$$is positive-definite, so the solution found above minimizes the objective function (the objective function is concave).

hbghlyj

kuing 发表于 2017-9-10 10:07
发现极值点有负数，可见最值只能在边 ...

$\nabla L=0$无解