问题2012060301

Show that the following inequalities hold for positive $a$, $b$, $c$.
(a) $a^2 + b^2 + c^2 \ge ab + bc + ca$.
(b) $(a + b)(b + c)(c + a) \ge 8abc$.
(c) $a^2b^2 + b^2c^2 + c^2a^2 \ge abc(a + b + c)$.

来源

Introduction to Calculus and Analysis

分析

这里的不等式都可以使用平均值不等式来证明.

解答

(a)
$$\begin{aligned}a^2 + b^2 + c^2 &= \frac{1}{2}(a^2 + b^2) + \frac{1}{2}(b^2 + c^2) + \frac{1}{2}(c^2 + a^2) \\&\ge ab + bc + ca\end{aligned}$$
(b)
$$\begin{aligned}(a + b)(b + c)(c + a) &\ge 2\sqrt{ab} \cdot 2\sqrt{bc} \cdot 2\sqrt{ca} \\& \ge 8abc\end{aligned}$$
(c)
$$\begin{aligned}a^2b^2 + b^2c^2 + c^2a^2 &= \frac{1}{2}(a^2b^2 + b^2c^2) + \frac{1}{2}(b^2c^2 + c^2a^2) + \frac{1}{2}(c^2a^2 + a^2b^2) \\&\ge abbc + bcca + caab \\ &= abc(a + b + c)\end{aligned}$$

问题2012060302

Assume that the numbers $x_1$, $x_2$, $x_3$ and $a_{ik}$($i, k = 1, 2, 3$) are all positive, and in addition, $a_{ik} \le M$ and $x_1^2 + x_2^2 + x_3^2 \le 1$. Prove that
$$a_{11}x_1^2 + a_{12}x_1x_2 + \cdots + a_{33}x_3^2 \le 3M.$$

来源

Introduction to Calculus and Analysis

解答

同样使用平均值不等式即可.
$$\begin{aligned}a_{11}x_1^2 + a_{12}x_1x_2 + \cdots + a_{33}x_3^2 &\le M(x_1^2 + x_1x_2 + \cdots + x_3^2) \\&= (x_1^2 + x_2^2 + x_3^2 + 2x_1x_2 + 2x_2x_3 + 2x_3x_1)M \\ &\le (x_1^2 + x_2^2 + x_3^2 + x_1^2 + x_2^2 +  x_2^2 + x_3^2 +  x_3^2 + x_1^2)M \\&= 3( x_1^2 + x_2^2 + x_3^2)M \\ &\le 3M\end{aligned}$$

问题2012060303

Prove the following inequality and give its geometrical interpretation for $n \le 3$,
$$\sqrt{(a_1 – b_1)^2 + \cdots + (a_n – b_n)^2} \le \sqrt{a_1^2 + \cdots + a_n^2} + \sqrt{b_1^2 + \cdots + b_n^2}.$$

来源

Introduction to Calculus and Analysis

解答

可以使用Cauchy不等式来证明. 我们使用分析法, 从需要证明的不等式出发, 从后面往前推.
$$\begin{aligned}&\ \sqrt{\sum_{i=1}^{n}{(a_i – b_i)^2}} \le \sqrt{\sum_{i=1}^{n}{a_i^2}} + \sqrt{\sum_{i=1}^{n}{b_i^2}} \\ &\Leftrightarrow \sum_{i=1}^{n}{(a_i – b_i)^2} \le (\sum_{i=1}^{n}{a_i^2}) + 2\sqrt{(\sum_{i=1}^{n}{a_i^2})(\sum_{i=1}^{n}{b_i^2})} + (\sum_{i=1}^{n}{(b_i^2)})  \\ &\Leftrightarrow-2\sum_{i=1}^{n}{(a_ib_i)} \le 2\sqrt{(\sum_{i=1}^{n}{a_i^2})(\sum_{i=1}^{n}{b_i^2})} \\ &\Leftrightarrow
-\sum_{i=1}^{n}{(a_ib_i)} \le \sqrt{(\sum_{i=1}^{n}{a_i^2})(\sum_{i=1}^{n}{b_i^2})}\end{aligned}$$
到了这一步, 我们基本上已经可以看出不等式了. 使用Cauchy不等式
$$\begin{aligned}&\ (\sum_{i=1}^{n}{(a_ib_i)})^2 \le (\sum_{i=1}^{n}{a_i^2})(\sum_{i=1}^{n}{b_i^2}) \\&\Leftrightarrow|\sum_{i=1}^{n}{(a_ib_i)}| \le \sqrt{(\sum_{i=1}^{n}{a_i^2})(\sum_{i=1}^{n}{b_i^2})}\end{aligned}$$
这就是我们要证明的不等式. 至于几何解释, 一句话: 两边之和大于第三边.

问题2012060304

Prove, and interpret geometrically for $n \le 3$,
$$\begin{aligned}&\sqrt{(a_1 + b_1 + \cdots + z_1)^2 + \cdots + (a_n + b_n + \cdots + z_n)^2} \\ &\ \le \sqrt{a_1^2 + \cdots + a_n^2} + \sqrt{b_1^2 + \cdots + b_n^2} + \cdots + \sqrt{z_1^2 + \cdots + z_n^2}.\end{aligned}$$

来源

Introduction to Calculus and Analysis

解答

使用上一题的结论(在上一题中把左边的负号换成正号也是成立的)来证明
$$\begin{aligned}&\sqrt{(a_1 + b_1 + \cdots + z_1)^2 + \cdots + (a_n + b_n + \cdots + z_n)^2} \\&\ \le \sqrt{a_1^2 + \cdots + a_n^2} + \sqrt{(b_1 + \cdots + z_1)^2 + \cdots + (b_n + \cdots + z_n)^2} \\&\ \le \sqrt{a_1^2 + \cdots + a_n^2} + \sqrt{b_1^2 + \cdots + b_n^2} + \sqrt{(c_1 + \cdots + z_1)^2 + \cdots + (c_n + \cdots + z_n)^2} \\&\ \le \sqrt{a_1^2 + \cdots + a_n^2} + \sqrt{b_1^2 + \cdots + b_n^2} + \cdots + \sqrt{z_1^2 + \cdots + z_n^2}\end{aligned}$$
至于几何解释, 我们考虑点:
$$\begin{aligned}&A(a_1+b_1+\cdots + z_1, \cdots, a_n+b_n+\cdots + z_n) \\&B(b_1+\cdots + z_1, \cdots, b_n+\cdots + z_n) \\&\cdots \\&Z(z_1, \cdots, z_n)
\end{aligned}$$
不等式转化为
$$|OA| \le |AB| + |BC| + \cdots + |ZO|,$$
几何意义还是: 两点之间直线最短.