腾讯课堂目标2017高中数学联赛基础班-2作业题解答-9

Posted 2020-08-22 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. 已知 $a_1, a_2, \cdots, a_n\in\mathbf{R^+}$, 满足 $a_1a_2\cdots a_n = 1$. 求证: $$(2 + a_1)(2 + a_2)\cdots(2 + a_n) \ge 3^n.$$ 解答: $$2 + a_i = 1+1+a_i \ge 3\sqrt[3]{a_i} \Rightarrow \sum_{i = 1}^{n} (2 + a_i) \ge 3^n\sqrt[3]{\prod_{i=1}^{n} a_i} = 3^n.$$

 

 

2. 设 $a, b, c\in\mathbf{R^+}$, 且 $a + b + c = 1$. 求证: $$(1 + a)(1 + b)(1 + c) \ge 8(1 - a)(1 - b)(1 - c).$$ 解答: $$\because\begin{cases}1 + a = (1 - b) + (1 - c) \ge 2\sqrt{(1-b)(1-c)}\\ 1 + b = (1 - a) + (1 - c) \ge 2\sqrt{(1 - a)(1 - c)}\\ 1 + c = (1 - a) + (1 - b) \ge 2\sqrt{(1 - a)(1 - b)}\end{cases}$$ $$\Rightarrow (1 + a)(1 + b)(1 + c) \ge 8\sqrt{(1-a)^2(1-b)^2(1-c)^2} = 8(1 - a)(1 - b)(1 - c).$$

 

 

3. 设 $a, b, c\in\mathbf{R^+}$, 满足 $a^2 + b^2 + c^2 = 1$. 求证: $${ab \over c} + {bc \over a} + {ca \over b} \ge \sqrt3.$$ 解答: $$\because\left({ab \over c} + {bc \over a} + {ca \over b}\right)^2 = {a^2b^2 \over c^2} + {b^2c^2 \over a^2} + {c^2a^2 \over b^2} + 2(a^2 + b^2 + c^2)$$ $$= {1\over2}\left({a^2b^2 \over c^2} + {b^2c^2 \over a^2}\right) + {1\over2}\left({a^2b^2 \over c^2} + {c^2a^2 \over b^2}\right) +{1\over2}\left({b^2c^2 \over a^2} + {c^2a^2 \over b^2}\right) + 2$$ $$\ge b^2 + a^2 + c^2 + 2 = 3$$ $$\therefore {ab \over c} + {bc \over a} + {ca \over b} \ge \sqrt3.$$

 

 

4. 设 $a$, $b$, $c$ 是非负数, 求证: $${1\over3}(a + b + c)^2 \ge a\sqrt{bc} + b\sqrt{ca} + c\sqrt{ab}.$$ 解答: $$\because(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$ $$= (a^2 + bc) + (b^2 + ca) + (c^2 + ab) + (ab + bc + ca)$$ $$\ge 2a\sqrt{bc} + 2b\sqrt{ca} + 2c\sqrt{ab} + (ab + bc + ca)$$ $$= 2a\sqrt{bc} + 2b\sqrt{ca} + 2c\sqrt{ab} + {1\over2}(ab + bc) + {1\over2}(bc + ca) + {1\over2}(ca + ab)$$ $$\ge 2a\sqrt{bc} + 2b\sqrt{ca} + 2c\sqrt{ab} + a\sqrt{bc} + b\sqrt{ca} + c\sqrt{ab}$$ $$= 3\left(a\sqrt{bc} + b\sqrt{ca} + c\sqrt{ab}\right).$$ $$\therefore {1\over3}(a + b + c)^2 \ge a\sqrt{bc} + b\sqrt{ca} + c\sqrt{ab}.$$

 

 

5. 设 $a, b, c\in\mathbf{R^+}$, 且 $(1 + a)(1 + b)(1 + c) = 8$. 求证: $abc \le 1$.

解答: $$8 = abc + ab + bc + ca + a + b + c + 1 \ge abc + 3\sqrt[3]{a^2b^2c^2} + 3\sqrt[3]{abc} + 1$$ $$= (1 + \sqrt[3]{abc})^3 \Rightarrow \sqrt[3]{abc} \le 1 \Rightarrow abc \le 1.$$

 

 

6. 设 $a, b, c\in\mathbf{R^+}$, 求证: $$\left(1 + {a\over b}\right)\left(1 + {b \over c}\right)\left(1 + {c \over a}\right) \ge 2\left(1 + {a + b + c \over \sqrt[3]{abc}}\right).$$ 解答: $$\left(1 + {a\over b}\right)\left(1 + {b \over c}\right)\left(1 + {c \over a}\right) = 1 + {a\over b} + {b \over c} + {c \over a} + {a \over c} + {c \over b} + {b \over a} + 1$$ $$= {a+b+c \over a} + {a+b+c \over b} + {a+b+c \over c} - 1$$ $$= (a + b + c)\left({1\over a} + {1\over b} + {1 \over c}\right) - 1 \ge (a + b + c)\cdot3\sqrt[3]{1\over abc} - 1$$ $$= {3(a+b+c) \over \sqrt[3]{abc}} - 1 = {2(a+b+c) \over \sqrt[3]{abc}} + {(a+b+c) \over \sqrt[3]{abc}} - 1$$ $$\ge {2(a+b+c) \over \sqrt[3]{abc}} + 3 - 1 = 2\left(1 + {a + b + c \over \sqrt[3]{abc}}\right).$$

 

 

7. 设 $a_1, a_2, \cdots, a_n, b_1, b_2, \cdots, b_n\in\mathbf{R^+}$, 且满足 $\displaystyle\sum_{k=1}^{n}a_k = \sum_{k = 1}^{n}b_k$. 求证: $$\sum_{k = 1}^{n}{a_k^2 \over a_k + b_k} \ge {1\over2}\sum_{k = 1}^{n}a_k.$$ 解答: $$\because {a_k^2 \over a_k + b_k} + {a_k + b_k \over 4} \ge a_k$$ $$\therefore \sum_{k = 1}^{n}{a_k^2 \over a_k + b_k} \ge \sum_{k = 1}^{n}a_k - {1\over4}\sum_{k = 1}^{n}(a_k + b_k) = {1\over2}\sum_{k = 1}^{n}a_k.$$

 

 

8. 设 $a_1, a_2, \cdots, a_n\in\mathbf{R^+}$, 满足 $a_1 + a_2 + \cdots + a_n = 1$. 求$$S = {a_1 \over 1 + a_2 + \cdots + a_n} + {a_2 \over 1 + a_1 + a_3 + \cdots + a_n} + \cdots + {a_n \over 1 + a_1 + \cdots + a_{n-1}}$$之最小值.

解答: $$S = {a_1 \over 2-a_1} + {a_2 \over 2-a_2} + \cdots + {a_n \over 2-a_n}$$ $$= {2 - (2 - a_1) \over 2-a_1} + {2 - (2 - a_2) \over 2-a_2} + \cdots + {2 - (2 - a_n) \over 2-a_n}$$ $$= 2\left({1\over 2-a_1} + {1 \over 2 - a_2} + \cdots + {1 \over 2 - a_n}\right) - n$$ $$\ge 2\cdot {n^2 \over (2-a_1) + (2 - a_2) + \cdots + (2 - a_n)} - n$$ $$= {2n^2 \over 2n - 1} - n = {n \over 2n-1}$$ 当且仅当 $a_1 = a_2 = \cdots = a_n = \displaystyle{1\over n}$ 时取等号.

上述不等号使用了 $H_n \le G_n$, 即 $${n \over {1\over 2-a_1} + {1 \over 2 - a_2} + \cdots + {1 \over 2 - a_n}} \le {(2-a_1) + (2 - a_2) + \cdots + (2 - a_n) \over n}$$ $$\Rightarrow {1\over 2-a_1} + {1 \over 2 - a_2} + \cdots + {1 \over 2 - a_n} \ge {n^2 \over (2-a_1) + (2 - a_2) + \cdots + (2 - a_n)}.$$