腾讯课堂目标2017高中数学联赛基础班-2作业题解答-12

 

1. 设 $x_i$ ($i = 1$, $2$, $\cdots$, $n$)均为非负实数, 且 $\sum x_i < \dfrac{1}{2}$, 证明: $$\prod_{i=1}^{n}(1 - x_i) > \frac{1}{2}.$$ 解答: $$\prod_{i=1}^{n}(1 - x_i) \ge (1 - x_1 - x_2)(1 - x_3)(1 - x_4)\cdots(1 - x_n)$$ $$\ge \left[1 - (x_1 + x_2 + x_3)\right](1 - x_4)\cdots(1 - x_n)$$ $$\ge\cdots\cdots\ge1 - (x_1 + x_2 + \cdots + x_n) > \frac{1}{2}.$$

 

 

2. 设 $0 < x < \dfrac{\pi}{2}$, 证明: $$\cos^2x + x\sin x < 2.$$ 解答: $$2 - \cos^2x - x\sin x = 2 - 1 + \sin^2x - x\sin x$$ $$= \sin^2x - x\sin x + 1$$ $$= (\sin x - 1)^2 + (2 - x)\sin x > 0.$$

 

 

3. 设 $a_i \in\mathbf{N^*}$, $a_i \ge 2$ ($i = 1$, $2$, $\cdots$, $n$)且互不相同, 证明: $$\prod_{i = 1}^{n}\left(1 - \frac{1}{a_i^2}\right) > \frac{1}{2}.$$ 解答: $$\prod_{i = 1}^{n}\left(1 - \frac{1}{a_i^2}\right) = \left(1 - \frac{1}{a_1^2}\right)\left(1 - \frac{1}{a_2^2}\right) \cdots \left(1 - \frac{1}{a_n^2}\right)$$ $$\ge \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\cdots\left(1 - \frac{1}{(n+1)^2}\right)$$ $$=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdots\frac{n}{n+1}\cdot\frac{n+2}{n+1} = \frac{1}{2}\cdot\frac{n+2}{n+1} > \frac{1}{2}.$$

 

 

4. 证明: $$\prod_{n = 1}^{50}\frac{2n-1}{2n} < \frac{1}{10}.$$ 解答: $$A = \prod_{n = 1}^{50}\frac{2n-1}{2n} = \frac{1}{2}\cdot\frac{3}{4}\cdots\frac{99}{100} < \frac{2}{3}\cdot\frac{4}{5}\cdots\frac{100}{101} = B$$ $$\Rightarrow A^2 < AB = \frac{1}{101} < \frac{1}{100} \Rightarrow A < \frac{1}{10}.$$

 

 

5. 设 $\alpha, \beta \in \left(0, \dfrac{\pi}{2}\right)$, 证明: $$\frac{1}{\cos^2\alpha} + \frac{1}{\sin^2\alpha \sin^2\beta\cos^2\beta} \ge 9.$$ 解答: $$\frac{1}{\cos^2\alpha} + \frac{1}{\sin^2\alpha \sin^2\beta\cos^2\beta} = \frac{1}{\cos^2\alpha} + \frac{4}{\sin^2\alpha\sin^22\beta}$$ $$\ge \frac{1}{\cos^2\alpha} + \frac{4}{\sin^2\alpha} = \sec^2\alpha + 4\csc^2\alpha = 1 + \tan^2\alpha + 4(1 + \cot^2\alpha)$$ $$= 5 + \tan^2\alpha + 4\cot^2\alpha \ge 5 + 2\sqrt4 = 9.$$

 

 

6. 设 $a, b, c \in[0, 1]$, 证明: $$\frac{a}{bc + 1} + \frac{b}{ca + 1} + \frac{c}{ab + 1} \le 2.$$ 解答:

$a = b = c = 0$ 时显然成立.

$a + b + c > 0$ 时, $$2a(bc + 1) - a(a + b + c) = a(bc + 1 - a) + a(bc - b - c + 1)$$ $$= a(bc + 1 - a) + a(b -1)(c - 1) \ge 0$$ $$\Rightarrow \frac{a}{bc + 1} \le \frac{2a}{a + b + c}$$ $$\Rightarrow \frac{a}{bc + 1} + \frac{b}{ca + 1} + \frac{c}{ab + 1} \le \frac{2a}{a + b + c} + \frac{2b}{a + b + c} + \frac{2c}{a + b + c} = 2.$$

 

 

7. $x, y \in \mathbf{R}$, 证明: $$\cos^2x + \cos^2y - \cos xy < 3.$$ 解答:

易知, $\cos^2x + \cos^2y - \cos xy \le 3$.

若 $\cos^2x + \cos^2y - \cos xy = 3$, 则 $$x^2 = 2k_1\pi,\ y^2 = 2k_2\pi,\ xy = (2k_3+1)\pi$$ $$\Rightarrow k_1k_2 = k_3^2 + k_3 + \frac{1}{4},\ (k_1, k_2, k_3\in\mathbf{Z}).$$ 矛盾. 因此 $\cos^2x + \cos^2y - \cos xy < 3$.

 

 

8. $a, b\in\mathbf{R^*}$, $\dfrac{1}{a} + \dfrac{1}{b} = 1$. 证明: 对一切正整数 $n$, 有 $$(a+b)^n - a^n - b^n \ge 2^{2n} - 2^{n+1}.$$ 解答:

$n=1$ 时, $(a+b) - a - b \ge 2^2 - 2^2$ 成立.

假设 $n = k$ 时成立, 即 $(a+b)^k - a^k - b^k \ge 2^{2k} - 2^{k+1}$, $$\Rightarrow (a + b)^{k+1} - a^{k+1} - b^{k+1} = (a + b)\left[(a+b)^k - a^k - b^k\right] + ab^k + a^kb$$ $$\because \frac{1}{a} + \frac{1}{b} = 1\Rightarrow ab = a+b$$ $$\because (a + b)\left(\frac{1}{a} + \frac{1}{b}\right) \ge 1 + 1 + 2 = 4 \Rightarrow ab = a+b \ge 4$$ $$\therefore (a + b)\left[(a+b)^k - a^k - b^k\right] + ab^k + a^kb \ge 4\left(2^{2k} - 2^{k+1}\right) + 2\sqrt{(ab)^{k+1}}$$ $$\ge 2^{2k+2} - 2^{k+3} + 2^{k+2} = 2^{2(k+1)} - 2^{k+2}.$$ 即 $n = k+1$ 时亦成立.