腾讯课堂目标2017高中数学联赛基础班-2作业题解答-10

Posted 2020-08-24 赵胤数学课

 

课程链接：目标2017高中数学联赛基础班-2（赵胤授课）

 

1. 设 $a_1, a_2, \cdots, a_n\in\mathbf{R}$, 证明: $$\sqrt[3]{a_1^3 + a_2^3 + \cdots + a_n^3} \le \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2}.$$ 解答: $$\left(\sum a_i^3\right)^2 \le \sum a_i^2 \cdot \sum a_i^4 \le \sum a_i^2 \cdot \left(\sum a_i^2\right)^2 = \left(\sum a_i^2\right)^3$$ 左右两边同时开 $6$ 次方即得证.

 

 

2. 设 $a_i, b_i, c_i, d_i \in\mathbf{R^{+}}$ ($i = 1, 2, \cdots, n$), 求证: $$\left(\sum_{i = 1}^{n}a_ib_ic_id_i\right)^4 \le \sum_{i = 1}^{n}a_i^4 \sum_{i = 1}^{n}b_i^4 \sum_{i = 1}^{n}c_i^4 \sum_{i = 1}^{n}d_i^4.$$ 解答: $$\left(\sum a_ib_ic_id_i\right)^4 \le \left[\sum\left(a_ib_i\right)^2 \cdot \sum\left(c_id_i\right)^2\right]^2$$ $$= \left(\sum a_i^2 b_i^2\right)^2 \cdot \left(\sum c_i^2d_i^2\right)^2 \le \sum a_i^4 \sum b_i^4 \sum c_i^4 \sum d_i^4.$$

 

 

3. 设 $x, y, z\in\mathbf{R}$, 求解下列方程 $$\begin{cases}2x + 3y + z = 13\\ 4x^2 + 9y^2 + z^2 - 2x + 15y + 3z = 82. \end{cases}$$ 解答: $$\begin{cases}2x + 3y + z = 13\\ 4x^2 + 9y^2 + z^2 - 2x + 15y + 3z = 82. \end{cases} \Rightarrow 4x^2 + 9y^2 + z^2 + 18y + 4z = 95$$ $$\Rightarrow (2x)^2 + (3y + 3)^2 + (z + 2)^2 = 108$$ 另一方面, $$2x + (3y + 3) + (z + 2) = 18$$ $$\Rightarrow 18^2 = \left[2x + (3y + 3) + (z + 2)\right]^2$$ $$\le 3\cdot\left((2x)^2 + (3y + 3)^2 + (z + 2)^2 \right) = 324$$ $$\Rightarrow 2x = 3y + 3 = z + 2 = 6$$ $$\Rightarrow \begin{cases}x = 3\\ y = 1\\ z = 4 \end{cases}$$

 

 

4. 设 $n\in\mathbf{N^{*}}$ 且 $n \ge 2$, 求证: $${4\over7} < 1 - {1\over2} + {1\over3} - {1\over4} + \cdots + {1 \over 2n - 1} - {1 \over 2n} < {\sqrt2 \over 2}.$$ 解答: $$1 - {1\over2} + {1\over3} - {1\over4} + \cdots + {1 \over 2n - 1} - {1 \over 2n}$$ $$= 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} - 2\cdot\left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n}\right)$$ $$= 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} - \left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right)$$ $$= \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}$$ 一方面 $$\left[(n+1) + (n+2) + \cdots + 2n\right]\cdot\left(\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}\right) > n^2$$ $$\Rightarrow \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}$$ $$ > \frac{2n}{3n+1} = \frac{2}{3}\left(1 - \frac{1}{3n+1}\right)\ge\frac{4}{7}.$$ 另一方面 $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}$$ $$ < n^{\frac{1}{2}}\cdot \left[\frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{(2n)^2}\right]^{\frac{1}{2}}$$ $$< \sqrt{n} \cdot \left[\frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} + \cdots + \frac{1}{(2n-1)2n}\right]^{\frac{1}{2}}$$ $$= \sqrt{n} \cdot \left(\frac{1}{n} - \frac{1}{2n}\right)^{\frac{1}{2}} = \sqrt{n}\cdot \sqrt{\frac{1}{2n}} = \frac{\sqrt2}{2}.$$ 综上, $${4\over7} < 1 - {1\over2} + {1\over3} - {1\over4} + \cdots + {1 \over 2n - 1} - {1 \over 2n} < {\sqrt2 \over 2}.$$

 

 

5. 设 $a, b, c, d\in\mathbf{N^{*}}$, 求证: $${a \over b+c} + {b \over c+d} + {c \over d+a} + {d \over a+b} \ge 2.$$ 解答: $$\left(\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}\right) \left[a(b+c) + b(c+d) + c(d + a) + d(a + b)\right]$$ $$\ge (a + b + c + d)^2$$ 另一方面 $$(a + b +c + d)^2 - 2\cdot\left[a(b+c) + b(c+d) + c(d + a) + d(a + b)\right]$$ $$= a^2 + b^2 + c^2 + d^2 - 2ac - 2bd = (a - c)^2 + (b - d)^2 \ge 0$$ 因此 $$\frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}$$ $$\ge \frac{(a + b +c + d)^2}{a(b+c) + b(c+d) + c(d + a) + d(a + b)} \ge 2.$$

 

 

6. 设 $a_i \in \mathbf{R^{+}}$ ($i = 1, 2, \cdots, n$), 且 $a_1 + a_2 + \cdots + a_n = 1$. 求证: $$\sum_{i = 1}^{n}\left(a_i + {1\over a_i}\right)^2 \ge {\left(n^2 + 1\right)^2 \over n}.$$ 解答: $$n\cdot\sum\left(a_i + \frac{1}{a_i}\right)^2 \ge \left[\sum\left(a_i + \frac{1}{a_i}\right)\right]^2 = \left(1 + \sum\frac{1}{a_i}\right)^2$$ $$ = \left(1 + \sum a_i \cdot\sum\frac{1}{a_i}\right)^2 \ge \left(1 + n^2\right)^2$$ $$\Rightarrow \sum_{i = 1}^{n}\left(a_i + {1\over a_i}\right)^2 \ge {\left(n^2 + 1\right)^2 \over n}.$$

 

 

7. 设 $a_i \in \mathbf{R^{+}}$ ($i = 1, 2, \cdots, n$), 求证: $${\left(a_1 + a_2 + \cdots + a_n\right)^2 \over 2\left(a_1^2 + a_2^2 + \cdots + a_n^2\right)} \le {a_1 \over a_2 + a_3} + {a_2 \over a_3 + a_4} + \cdots + {a_n \over a_1 + a_2}.$$ 解答:

对右式表达成通项并利用Cauchy不等式去分母: $$\sum\frac{a_i}{a_{i+1} + a_{i+2}} \cdot \sum a_i\left(a_{i+1} + a_{i+2}\right) \ge \left(\sum a_i\right)^2$$ $$\Rightarrow \frac{\left(\sum a_i\right)^2}{\sum\dfrac{a_i}{a_{i+1} + a_{i+2}}} \le \sum a_i\left(a_{i+1} + a_{i+2}\right)$$ 因此仅需证明 $$\sum a_i\left(a_{i+1} + a_{i+2}\right) \le 2\cdot\sum a_i^2$$ 可用A-G不等式得出: $$\sum a_i\left(a_{i+1} + a_{i+2}\right) \le \frac{1}{2}\cdot\sum\left[\left(a_i^2 + a_{i+1}^2\right) + \left(a_i^2 + a_{i+2}^2\right)\right]$$ $$= \frac{1}{2}\cdot\left(a_1^2 + a_2^2 + a_1^2 + a_3^2 + a_2^2 + a_3^2 + a_2^2 + a_4^2 + \cdots + a_n^2 + a_1^2 + a_n^2 + a_2^2\right)$$ $$= 2\left(a_1^2 + a_2^2 + \cdots + a_n^2\right) = 2\cdot\sum a_i^2.$$

 

 

8. 对于满足 $1 \le r \le s \le t \le 4$ 的一切实数 $r$, $s$, $t$, 求下式的最小值: $$(r - 1)^2 + \left({s\over r} - 1\right)^2 + \left({t \over s} - 1\right)^2 + \left({4 \over t} - 1\right)^2.$$ 解答:

注意到 $r$, $\dfrac{s}{r}$, $\dfrac{t}{s}$, $\dfrac{4}{t}$ 乘积为定值, 可使用A-G不等式求最小值.

因此考虑使用Cauchy不等式构造它们的和: $$4\cdot \left[(r - 1)^2 + \left({s\over r} - 1\right)^2 + \left({t \over s} - 1\right)^2 + \left({4 \over t} - 1\right)^2\right]$$ $$\ge \left[(r - 1) + \left(\frac{s}{r} - 1\right) + \left(\frac{t}{s} - 1\right) + \left(\frac{4}{t} - 1\right)\right]^2$$ $$= \left(r + \frac{s}{r} + \frac{t}{s} + \frac{4}{t} - 4\right)^2 \ge \left(4\sqrt[4]{4} - 4\right)^2 = 16\left(\sqrt2 - 1\right)^2$$ 当 $r = \sqrt2$, $s = 2$, $t = 2\sqrt2$ 时取等号. 其最小值为 $4\left(\sqrt2 - 1\right)^2$.