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[不等式] 证明双曲集是凸的

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hbghlyj Posted at 2022-10-21 05:50:35 |Read mode
hw1
Hyperbolic sets. 证明双曲集 $\left\{x \in \mathbf{R}_{+}^2 \mid x_1 x_2 \geq 1\right\}$ 是凸的. 一般地, 证明 $\left\{x \in \mathbf{R}_{+}^n \mid \prod_{i=1}^n x_i \geq 1\right\}$ 是凸的. 提示. 若 $a, b \geq 0$ 且 $0 \leq \theta \leq 1$, 则 $a^\theta b^{1-\theta} \leq \theta a+(1-\theta) b$; 见 §3.1.9.
解.
(a) 我们在不使用提示的情况下证明了第一部分。 考虑这个集合中两点 $\left(x_1, x_2\right)$ 与 $\left(y_1, y_2\right)$ 的凸组合 $z$. 若 $x \succeq y$, 则 $z=\theta x+(1-\theta) y \succeq y$, 且显然 $z_1 z_2 \geq y_1 y_2 \geq 1$. 可类似对 $y \succeq x$ 证明.
假设 $y \nsucceq x$ 与 $x \nsucceq y$, 即, $\left(y_1-x_1\right)\left(y_2-x_2\right)<0$. 则
\[
\begin{aligned}
&\left(\theta x_1+(1-\theta) y_1\right)\left(\theta x_2+(1-\theta) y_2\right) \\
&\quad=\theta^2 x_1 x_2+(1-\theta)^2 y_1 y_2+\theta(1-\theta) x_1 y_2+\theta(1-\theta) x_2 y_1 \\
&\quad=\theta x_1 x_2+(1-\theta) y_1 y_2-\theta(1-\theta)\left(y_1-x_1\right)\left(y_2-x_2\right) \\
&\quad \geq 1 .
\end{aligned}
\]
(b) 设 $\prod_i x_i \geq 1$ 与 $\prod_i y_i \geq 1$. 使用提示中的不等式,我们有
\[
\prod_i\left(\theta x_i+(1-\theta) y_i\right) \geq \prod x_i^\theta y_i^{1-\theta}=\left(\prod_i x_i\right)^\theta\left(\prod_i y_i\right)^{1-\theta} \geq 1
\]

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 Author| hbghlyj Posted at 2022-10-21 06:33:01
$a^\theta b^{1-\theta} \leq \theta a+(1-\theta) b$

加权平均值
Weighted AM-GM Inequality

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2025-4-21 19:02 GMT+8

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