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: $$prod_{i=1}^{n}(1 - x_i) ge (1 - x_1 - x_2)(1 - x_3)(1 - x_4)cdots(1 - x_n)$$ $$ge left[1 - (x_1 + x_2 + x_3) ight](1 - x_4)cdots(1 - x_n)$$ $$gecdotscdotsge1 - (x_1 + x_2 + cdots + x_n) > frac{1}{2}.$$
2. 设 $0 < x < dfrac{pi}{2}$, 证明: $$cos^2x + xsin x < 2.$$ 解答: $$2 - cos^2x - xsin x = 2 - 1 + sin^2x - xsin x$$ $$= sin^2x - xsin x + 1$$ $$= (sin x - 1)^2 + (2 - x)sin x > 0.$$
3. 设 $a_i inmathbf{N^*}$, $a_i ge 2$ ($i = 1$, $2$, $cdots$, $n$)且互不相同, 证明: $$prod_{i = 1}^{n}left(1 - frac{1}{a_i^2} ight) > frac{1}{2}.$$ 解答: $$prod_{i = 1}^{n}left(1 - frac{1}{a_i^2} ight) = left(1 - frac{1}{a_1^2} ight)left(1 - frac{1}{a_2^2} ight) cdots left(1 - frac{1}{a_n^2} ight)$$ $$ge left(1 - frac{1}{2^2} ight)left(1 - frac{1}{3^2} ight)cdotsleft(1 - frac{1}{(n+1)^2} ight)$$ $$=frac{1}{2}cdotfrac{3}{2}cdotfrac{2}{3}cdotfrac{4}{3}cdotsfrac{n}{n+1}cdotfrac{n+2}{n+1} = frac{1}{2}cdotfrac{n+2}{n+1} > frac{1}{2}.$$
4. 证明: $$prod_{n = 1}^{50}frac{2n-1}{2n} < frac{1}{10}.$$ 解答: $$A = prod_{n = 1}^{50}frac{2n-1}{2n} = frac{1}{2}cdotfrac{3}{4}cdotsfrac{99}{100} < frac{2}{3}cdotfrac{4}{5}cdotsfrac{100}{101} = B$$ $$Rightarrow A^2 < AB = frac{1}{101} < frac{1}{100} Rightarrow A < frac{1}{10}.$$
5. 设 $alpha, eta in left(0, dfrac{pi}{2} ight)$, 证明: $$frac{1}{cos^2alpha} + frac{1}{sin^2alpha sin^2etacos^2eta} ge 9.$$ 解答: $$frac{1}{cos^2alpha} + frac{1}{sin^2alpha sin^2etacos^2eta} = frac{1}{cos^2alpha} + frac{4}{sin^2alphasin^22eta}$$ $$ge frac{1}{cos^2alpha} + frac{4}{sin^2alpha} = sec^2alpha + 4csc^2alpha = 1 + an^2alpha + 4(1 + cot^2alpha)$$ $$= 5 + an^2alpha + 4cot^2alpha ge 5 + 2sqrt4 = 9.$$
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 $$x^2 = 2k_1pi, y^2 = 2k_2pi, xy = (2k_3+1)pi$$ $$Rightarrow k_1k_2 = k_3^2 + k_3 + frac{1}{4}, (k_1, k_2, k_3inmathbf{Z}).$$ 矛盾. 因此 $cos^2x + cos^2y - cos xy < 3$.
8. $a, binmathbf{R^*}$, $dfrac{1}{a} + dfrac{1}{b} = 1$. 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 $(a+b)^k - a^k - b^k ge 2^{2k} - 2^{k+1}$, $$Rightarrow (a + b)^{k+1} - a^{k+1} - b^{k+1} = (a + b)left[(a+b)^k - a^k - b^k ight] + ab^k + a^kb$$ $$ecause frac{1}{a} + frac{1}{b} = 1Rightarrow ab = a+b$$ $$ecause (a + b)left(frac{1}{a} + frac{1}{b} ight) ge 1 + 1 + 2 = 4 Rightarrow ab = a+b ge 4$$ $$ herefore (a + b)left[(a+b)^k - a^k - b^k ight] + ab^k + a^kb ge 4left(2^{2k} - 2^{k+1} ight) + 2sqrt{(ab)^{k+1}}$$ $$ge 2^{2k+2} - 2^{k+3} + 2^{k+2} = 2^{2(k+1)} - 2^{k+2}.$$ 即 $n = k+1$ 时亦成立.