1. 设 $a_1, a_2, cdots, a_ninmathbf{R}$, 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: $$left(sum a_i^3 ight)^2 le sum a_i^2 cdot sum a_i^4 le sum a_i^2 cdot left(sum a_i^2 ight)^2 = left(sum a_i^2 ight)^3$$ 左右两边同时开 $6$ 次方即得证.
2. 设 $a_i, b_i, c_i, d_i inmathbf{R^{+}}$ ($i = 1, 2, cdots, n$), 求证: $$left(sum_{i = 1}^{n}a_ib_ic_id_i ight)^4 le sum_{i = 1}^{n}a_i^4 sum_{i = 1}^{n}b_i^4 sum_{i = 1}^{n}c_i^4 sum_{i = 1}^{n}d_i^4.$$ 解答: $$left(sum a_ib_ic_id_i ight)^4 le left[sumleft(a_ib_i ight)^2 cdot sumleft(c_id_i ight)^2 ight]^2$$ $$= left(sum a_i^2 b_i^2 ight)^2 cdot left(sum c_i^2d_i^2 ight)^2 le sum a_i^4 sum b_i^4 sum c_i^4 sum d_i^4.$$
3. 设 $x, y, zinmathbf{R}$, 求解下列方程 $$egin{cases}2x + 3y + z = 13\ 4x^2 + 9y^2 + z^2 - 2x + 15y + 3z = 82. end{cases}$$ 解答: $$egin{cases}2x + 3y + z = 13\ 4x^2 + 9y^2 + z^2 - 2x + 15y + 3z = 82. end{cases} Rightarrow 4x^2 + 9y^2 + z^2 + 18y + 4z = 95$$ $$Rightarrow (2x)^2 + (3y + 3)^2 + (z + 2)^2 = 108$$ 另一方面, $$2x + (3y + 3) + (z + 2) = 18$$ $$Rightarrow 18^2 = left[2x + (3y + 3) + (z + 2) ight]^2$$ $$le 3cdotleft((2x)^2 + (3y + 3)^2 + (z + 2)^2 ight) = 324$$ $$Rightarrow 2x = 3y + 3 = z + 2 = 6$$ $$Rightarrow egin{cases}x = 3\ y = 1\ z = 4 end{cases}$$
4. 设 $ninmathbf{N^{*}}$ 且 $n ge 2$, 求证: $${4over7} < 1 - {1over2} + {1over3} - {1over4} + cdots + {1 over 2n - 1} - {1 over 2n} < {sqrt2 over 2}.$$ 解答: $$1 - {1over2} + {1over3} - {1over4} + cdots + {1 over 2n - 1} - {1 over 2n}$$ $$= 1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{2n} - 2cdotleft(frac{1}{2} + frac{1}{4} + cdots + frac{1}{2n} ight)$$ $$= 1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{2n} - left(1 + frac{1}{2} + cdots + frac{1}{n} ight)$$ $$= frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{2n}$$ 一方面 $$left[(n+1) + (n+2) + cdots + 2n ight]cdotleft(frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{2n} ight) > n^2$$ $$Rightarrow frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{2n}$$ $$ > frac{2n}{3n+1} = frac{2}{3}left(1 - frac{1}{3n+1} ight)gefrac{4}{7}.$$ 另一方面 $$frac{1}{n+1} + frac{1}{n+2} + cdots + frac{1}{2n}$$ $$ < n^{frac{1}{2}}cdot left[frac{1}{(n+1)^2} + frac{1}{(n+2)^2} + cdots + frac{1}{(2n)^2} ight]^{frac{1}{2}}$$ $$< sqrt{n} cdot left[frac{1}{n(n+1)} + frac{1}{(n+1)(n+2)} + cdots + frac{1}{(2n-1)2n} ight]^{frac{1}{2}}$$ $$= sqrt{n} cdot left(frac{1}{n} - frac{1}{2n} ight)^{frac{1}{2}} = sqrt{n}cdot sqrt{frac{1}{2n}} = frac{sqrt2}{2}.$$ 综上, $${4over7} < 1 - {1over2} + {1over3} - {1over4} + cdots + {1 over 2n - 1} - {1 over 2n} < {sqrt2 over 2}.$$
5. 设 $a, b, c, dinmathbf{N^{*}}$, 求证: $${a over b+c} + {b over c+d} + {c over d+a} + {d over a+b} ge 2.$$ 解答: $$left(frac{a}{b+c} + frac{b}{c+d} + frac{c}{d+a} + frac{d}{a+b} ight) left[a(b+c) + b(c+d) + c(d + a) + d(a + b) ight]$$ $$ge (a + b + c + d)^2$$ 另一方面 $$(a + b +c + d)^2 - 2cdotleft[a(b+c) + b(c+d) + c(d + a) + d(a + b) ight]$$ $$= a^2 + b^2 + c^2 + d^2 - 2ac - 2bd = (a - c)^2 + (b - d)^2 ge 0$$ 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: $$sum_{i = 1}^{n}left(a_i + {1over a_i} ight)^2 ge {left(n^2 + 1 ight)^2 over n}.$$ 解答: $$ncdotsumleft(a_i + frac{1}{a_i} ight)^2 ge left[sumleft(a_i + frac{1}{a_i} ight) ight]^2 = left(1 + sumfrac{1}{a_i} ight)^2$$ $$ = left(1 + sum a_i cdotsumfrac{1}{a_i} ight)^2 ge left(1 + n^2 ight)^2$$ $$Rightarrow sum_{i = 1}^{n}left(a_i + {1over a_i} ight)^2 ge {left(n^2 + 1 ight)^2 over n}.$$
7. 设 $a_i in mathbf{R^{+}}$ ($i = 1, 2, cdots, n$), 求证: $${left(a_1 + a_2 + cdots + a_n ight)^2 over 2left(a_1^2 + a_2^2 + cdots + a_n^2 ight)} le {a_1 over a_2 + a_3} + {a_2 over a_3 + a_4} + cdots + {a_n over a_1 + a_2}.$$ 解答:
对右式表达成通项并利用Cauchy不等式去分母: $$sumfrac{a_i}{a_{i+1} + a_{i+2}} cdot sum a_ileft(a_{i+1} + a_{i+2} ight) ge left(sum a_i ight)^2$$ $$Rightarrow frac{left(sum a_i ight)^2}{sumdfrac{a_i}{a_{i+1} + a_{i+2}}} le sum a_ileft(a_{i+1} + a_{i+2} ight)$$ 因此仅需证明 $$sum a_ileft(a_{i+1} + a_{i+2} ight) le 2cdotsum a_i^2$$ 可用A-G不等式得出: $$sum a_ileft(a_{i+1} + a_{i+2} ight) le frac{1}{2}cdotsumleft[left(a_i^2 + a_{i+1}^2 ight) + left(a_i^2 + a_{i+2}^2 ight) ight]$$ $$= frac{1}{2}cdotleft(a_1^2 + a_2^2 + a_1^2 + a_3^2 + a_2^2 + a_3^2 + a_2^2 + a_4^2 + cdots + a_n^2 + a_1^2 + a_n^2 + a_2^2 ight)$$ $$= 2left(a_1^2 + a_2^2 + cdots + a_n^2 ight) = 2cdotsum a_i^2.$$
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: $$(r - 1)^2 + left({sover r} - 1 ight)^2 + left({t over s} - 1 ight)^2 + left({4 over t} - 1 ight)^2.$$ 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: $$4cdot left[(r - 1)^2 + left({sover r} - 1 ight)^2 + left({t over s} - 1 ight)^2 + left({4 over t} - 1 ight)^2 ight]$$ $$ge left[(r - 1) + left(frac{s}{r} - 1 ight) + left(frac{t}{s} - 1 ight) + left(frac{4}{t} - 1 ight) ight]^2$$ $$= left(r + frac{s}{r} + frac{t}{s} + frac{4}{t} - 4 ight)^2 ge left(4sqrt[4]{4} - 4 ight)^2 = 16left(sqrt2 - 1 ight)^2$$ 当 $r = sqrt2$, $s = 2$, $t = 2sqrt2$ 时取等号. 其最小值为 $4left(sqrt2 - 1 ight)^2$.