1. 已知 $a_1, a_2, cdots, a_ninmathbf{R^+}$, 满足 $a_1a_2cdots a_n = 1$. 求证: $$(2 + a_1)(2 + a_2)cdots(2 + a_n) ge 3^n.$$ 解答: $$2 + a_i = 1+1+a_i ge 3sqrt[3]{a_i} Rightarrow sum_{i = 1}^{n} (2 + a_i) ge 3^nsqrt[3]{prod_{i=1}^{n} a_i} = 3^n.$$
2. 设 $a, b, cinmathbf{R^+}$, 且 $a + b + c = 1$. 求证: $$(1 + a)(1 + b)(1 + c) ge 8(1 - a)(1 - b)(1 - c).$$ 解答: $$ecauseegin{cases}1 + a = (1 - b) + (1 - c) ge 2sqrt{(1-b)(1-c)}\ 1 + b = (1 - a) + (1 - c) ge 2sqrt{(1 - a)(1 - c)}\ 1 + c = (1 - a) + (1 - b) ge 2sqrt{(1 - a)(1 - b)}end{cases}$$ $$Rightarrow (1 + a)(1 + b)(1 + c) ge 8sqrt{(1-a)^2(1-b)^2(1-c)^2} = 8(1 - a)(1 - b)(1 - c).$$
3. 设 $a, b, cinmathbf{R^+}$, 满足 $a^2 + b^2 + c^2 = 1$. 求证: $${ab over c} + {bc over a} + {ca over b} ge sqrt3.$$ 解答: $$ecauseleft({ab over c} + {bc over a} + {ca over b} ight)^2 = {a^2b^2 over c^2} + {b^2c^2 over a^2} + {c^2a^2 over b^2} + 2(a^2 + b^2 + c^2)$$ $$= {1over2}left({a^2b^2 over c^2} + {b^2c^2 over a^2} ight) + {1over2}left({a^2b^2 over c^2} + {c^2a^2 over b^2} ight) +{1over2}left({b^2c^2 over a^2} + {c^2a^2 over b^2} ight) + 2$$ $$ge b^2 + a^2 + c^2 + 2 = 3$$ $$ herefore {ab over c} + {bc over a} + {ca over b} ge sqrt3.$$
4. 设 $a$, $b$, $c$ 是非负数, 求证: $${1over3}(a + b + c)^2 ge asqrt{bc} + bsqrt{ca} + csqrt{ab}.$$ 解答: $$ecause(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$ $$= (a^2 + bc) + (b^2 + ca) + (c^2 + ab) + (ab + bc + ca)$$ $$ge 2asqrt{bc} + 2bsqrt{ca} + 2csqrt{ab} + (ab + bc + ca)$$ $$= 2asqrt{bc} + 2bsqrt{ca} + 2csqrt{ab} + {1over2}(ab + bc) + {1over2}(bc + ca) + {1over2}(ca + ab)$$ $$ge 2asqrt{bc} + 2bsqrt{ca} + 2csqrt{ab} + asqrt{bc} + bsqrt{ca} + csqrt{ab}$$ $$= 3left(asqrt{bc} + bsqrt{ca} + csqrt{ab} ight).$$ $$ herefore {1over3}(a + b + c)^2 ge asqrt{bc} + bsqrt{ca} + csqrt{ab}.$$
5. 设 $a, b, cinmathbf{R^+}$, 且 $(1 + a)(1 + b)(1 + c) = 8$. 求证: $abc le 1$.
解答: $$8 = abc + ab + bc + ca + a + b + c + 1 ge abc + 3sqrt[3]{a^2b^2c^2} + 3sqrt[3]{abc} + 1$$ $$= (1 + sqrt[3]{abc})^3 Rightarrow sqrt[3]{abc} le 1 Rightarrow abc le 1.$$
6. 设 $a, b, cinmathbf{R^+}$, 求证: $$left(1 + {aover b} ight)left(1 + {b over c} ight)left(1 + {c over a} ight) ge 2left(1 + {a + b + c over sqrt[3]{abc}} ight).$$ 解答: $$left(1 + {aover b} ight)left(1 + {b over c} ight)left(1 + {c over a} ight) = 1 + {aover b} + {b over c} + {c over a} + {a over c} + {c over b} + {b over a} + 1$$ $$= {a+b+c over a} + {a+b+c over b} + {a+b+c over c} - 1$$ $$= (a + b + c)left({1over a} + {1over b} + {1 over c} ight) - 1 ge (a + b + c)cdot3sqrt[3]{1over abc} - 1$$ $$= {3(a+b+c) over sqrt[3]{abc}} - 1 = {2(a+b+c) over sqrt[3]{abc}} + {(a+b+c) over sqrt[3]{abc}} - 1$$ $$ge {2(a+b+c) over sqrt[3]{abc}} + 3 - 1 = 2left(1 + {a + b + c over sqrt[3]{abc}} ight).$$
7. 设 $a_1, a_2, cdots, a_n, b_1, b_2, cdots, b_ninmathbf{R^+}$, 且满足 $displaystylesum_{k=1}^{n}a_k = sum_{k = 1}^{n}b_k$. 求证: $$sum_{k = 1}^{n}{a_k^2 over a_k + b_k} ge {1over2}sum_{k = 1}^{n}a_k.$$ 解答: $$ecause {a_k^2 over a_k + b_k} + {a_k + b_k over 4} ge a_k$$ $$ herefore sum_{k = 1}^{n}{a_k^2 over a_k + b_k} ge sum_{k = 1}^{n}a_k - {1over4}sum_{k = 1}^{n}(a_k + b_k) = {1over2}sum_{k = 1}^{n}a_k.$$
8. 设 $a_1, a_2, cdots, a_ninmathbf{R^+}$, 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: $$S = {a_1 over 2-a_1} + {a_2 over 2-a_2} + cdots + {a_n over 2-a_n}$$ $$= {2 - (2 - a_1) over 2-a_1} + {2 - (2 - a_2) over 2-a_2} + cdots + {2 - (2 - a_n) over 2-a_n}$$ $$= 2left({1over 2-a_1} + {1 over 2 - a_2} + cdots + {1 over 2 - a_n} ight) - n$$ $$ge 2cdot {n^2 over (2-a_1) + (2 - a_2) + cdots + (2 - a_n)} - n$$ $$= {2n^2 over 2n - 1} - n = {n over 2n-1}$$ 当且仅当 $a_1 = a_2 = cdots = a_n = displaystyle{1over n}$ 时取等号.
上述不等号使用了 $H_n le G_n$, 即 $${n over {1over 2-a_1} + {1 over 2 - a_2} + cdots + {1 over 2 - a_n}} le {(2-a_1) + (2 - a_2) + cdots + (2 - a_n) over n}$$ $$Rightarrow {1over 2-a_1} + {1 over 2 - a_2} + cdots + {1 over 2 - a_n} ge {n^2 over (2-a_1) + (2 - a_2) + cdots + (2 - a_n)}.$$