考研数学三 经济ch10 各章复习题目及答案

第十章 函数方程与不等式证明

an?1an?a?一. 证明不等式

(n?1)2lna证明: 令f(x)?ax, 在1111n?1an?2. (a > 1, n ? 1) n1?n?1,1n?上使用拉格朗日定理

f(1)?f(1)?f'(?)(1?1)nn?1nn?1 1 11?nn?1a?a?alnan(n?1) 即

a1n?alna1n?1a? ?n(n?1)111 所以

an?1an?a?

(n?1)2lna

二. 若a ? 0, b ? 0, 0 < p < 1, 证明 (a?b)p?ap?bp

1n?1an?2. (a > 1, n ? 1) n证明: 令f(x)?(x?b)p?xp?bp 显然f(0) = 0. 当x ? 0 时, 因为0 < p < 1

f'(x)?p(x?b)p?1?pxp?1?0

所以当x ? 0时, f(x)单减, 所以f(a) ? f(0) = 0. 所以 (a?b)?a?b?0 即得 (a?b)?a?b

三. 设函数f(x)在[0, 1]上有连续导数, 满足0?f'(x)?1且f(0)?0. 求证

113??f(x)dx?f(x)dx ????00??2ppppppxx3证明: 令F(x)????0f(t)dt????0f(t)dt, 显然F(0) = 0. 因为0?f'(x)?1且f(0)?0,

??2所以当x > 0时f(x) > 0. F'(x)?2f(x)?x0f(t)dt?f3(x)

1

=f(x)??2??x0f(t)dt?f2(x)?? (1)

?令?(x)?2?x0f(t)dt?f2(x), 显然?(0) = 0.

?'(x)?2f(x)?2f(x)f'(x)?2f(x)(1?f'(x))?0

所以当x > 0时, ?(x) > 0. 由(1)知F'(x)?0(x > 0). 当x > 0时F(x) ? F(0) = 0.所以F(1) ? F(0) = 0. 立即得到

113 ??f(x)dx???f(x)dx

??0?0?2

四. 求证 |a|p?|b|p?21?p(|a|?|b|)p, (0 < p < 1). 求证: 先证当0 ? x ? 1, 0 < p < 1时, 有 21?p?xp?(1?x)p?1 令F(x)?xp?(1?x)p

F'(x)?pxp?1?p(1?x)p?1. F'(x)?0得 x?12. F(12)?21?p，F(1)?F(0)?1.

所以F(1)?221?p为最大值，F(1)?F(0)?1为最小值. 所以当0 ? x ? 1, 0 < p < 1时,

有

21?p?xp?(1?x)p?1

|b||a|, 则1?x?. 代入上述结论, 立即得到

|a|?|b||a|?|b|2? 令x? 21?p|a|p|b|p???1 pp(|a|?|b|)(|a|?|b|)ppp1?p即 (|a|?|b|)?|a|?|b|?2

(|a|?|b|)p, (0 < p < 1).

五. 求证: 若x + y + z = 6, 则x?y?z?12, (x ? 0, y ? 0, z ? 0). 证明：方法1:

2(x?y?z)?2xy?2yz?2xz

222222x2?y2?z2?(x?y?z)2?2xy?2yz?2xz?36?2(x?y?z)222

2

所以 3(x2?y2?z2)?36, x2?y2?z2?12 方法2:

?s(x,y,z)?x2?y2?z2解以下条件极值问题:?

?条件：x?y?z?6令F(x, y, z, ?) = x2 + y2 + z2－?(x + y + z－6)

Fx'?2x???0, Fy'?2y???0, Fz'?2z???0

解得 x = y = z = 2. 只有一个驻点, 当x = y = z = 2时达到最小值12. 所以 x2?y2?z2?12, (x ? 0, y ? 0, z ? 0)

六. 证明: 1? 若f(x)在[a, b]上是增加的，且在其上f''(x)?0，则 (b?a)f(a)??baf(x)dx?(b?a)f(a)?f(b)

22? 若f(x)在[a, b]上是增加的，且在其上f''(x)?0，则 (b?a)f(b)??baf(x)dx?(b?a)f(a)?f(b)

2证明：1? 方法1: 因为f(x)是增加的, 所以对于[a, b]中的一切x, 有f(x) > f(a), 所以 令F(x)??xbaf(x)dx?f(a)(b?a)

f(a)?f(x) ?a2f(a)?f(x)f'(x)f(x)?f(a)f'(x)?(x?a)??(x?a) F'(x)?f(x)?

2222f'(?)(x?a)f'(x)?(x?a) (a???x) ?221 =(x?a)[f'(?)?f'(x)]?0(因为f''(x)?0)

2f(t)dt?(x?a)所以F(x)单增. 又因为F(a) = 0, 所以F(b) > F(a) = 0. 立即可得

?baf(x)dx?(b?a)f(a)?f(b)

2方法2: 将f(x)台劳展开

f''(?)(t?x)2 2!f''(?1)(a?x)2 所以 f(a)?f(x)?f'(x)(a?x)?2!f''(?2)(b?x)2 f(b)?f(x)?f'(x)(b?x)?2!f''(?1)f''(?2)(a?x)2?(b?x)2 f(a)?f(b)?2f(x)?f'(x)(a?b)?2xf'(x)?2!2!?t, x, f(t)?f(x)?f'(x)(t?x)?

3

上式二边积分得 (f(b)?f(a))(b?a)?2 ?2所以

?baf(x)dx?(a?b)?f'(x)dx

ab?bab?f''(?)f''(?2)?1xf'(x)dx???(a?x)2?(b?x)2?dx

a2!?2!?b(f(b)?f(a))(b?a)?2?f(x)dx?(a?b)(f(b)?f(a))?2xf(x)?2?f(x)dx

aaa ?4 ?4bb?baf(x)dx?af(b)?af(a)?bf(b)?bf(a)?2bf(b)?2af(a)

?baf(x)dx?(b?a)(f(a)?f(b))

于是 2(b?a)(f(b)?f(a))?4即

?baf(x)dx

bb?a(f(b)?f(a))??f(x)dx

a22? 证法同1?.

注: 无论方法1?, 2?, 右边的不等式都不需要f(x)单增的条件.

x?x2???xn七. 证明: 1? 1?n2?

22x12?x2???xn

nx1?x2???xnn?x1x2?xn

n2nn?n?22证明: 1?方法一: 先证??akbk???ak?bk

k?1k?1?k?1?n?n2?2?n?2由 ?(akx?bk)???ak?x?2??akbk?x??bk?0

k?1k?1?k?1??k?1?2nn?n?22得到 ??akbk???ak?bk

k?1k?1?k?1?2n上述不等式中令ak?xk，bk?1, 得到 n???xk ?k?1?n??n??n??x2?n?1

?k?nk?1??22x12?x2???xn.

n2x?x2???xn即 1?n

4

方法二: 令f(x)?x2, p1?p2???pn?因为 f''(x)?2?0

1 n所以 f(p1x1??pnxn)?p1f(x1)???pnf(xn)

2x1???xn2x12???xn)?即 (

nnx?x2???xn即 1?n2? 取 f(x) = ln x, f''(x)??所以 ln?22x12?x2???xn

n1?0. 令p1 = p2 = … = pn = 1/n. x21?x1???xn?1?lnx???lnxn?lnnx1?xn ?1nn??nx1?x2???xnn?x1x2?xn.

n立即得到

八. 设f''(x)?c[a,b], 且f(a)?f(b)?0, 求证:

b

?a(b?a)3f(x)dx?max|f''(x)|

12a?x?b1b1b证明: 方法1: ?f''(x)(x?a)(x?b)dx??(x?a)(x?b)df'(x)

2a2a=

11(x?a)(x?b)f'(x)??f'(x)(2x?a?b)dx 22abb

=?111b(2x?a?b)df(x)??f(x)(2x?a?b)?f(x)2dx ??aa222ab =

?babf(x)dx

1b1bf(x)dx| =?f''(x)(x?a)(x?b)dx??|f''(x)(x?a)(x?b)|dx

2a2a所以 |

?ab1(b?a)3max|f''(x)| ?max|f''(x)|?(x?a)(b?x)dx=

a2a?x?b12a?x?b 5

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