讲高数17堂专题课不等式问题

1）单调性 5）凹凸性【证】f(xx21xln21xf(0)f(x)2xln2(1x)2ln(1x)，f(0)f(x)22ln(1x)1

1 1

[xln(1x)] xx1

ln(1x) 2】x0(1x)xe2(11)ln(1x)1x 2(1x)ln(1x)2x令f(x)2xx22(1x)ln(1 (x3】设0aba2

lnblna b 日中值定理知，至少存在一点(a,b)，lnblna(lnx)ba

1由于0ab，故11

a2lnblnab

.a2x(x)lnxlna (xx(x)1x

1a a2

(x a 2x

xa时,(x)单调减少.又(a)0xa时(x)(a)0lnxlnaxa从而当ba时，lnblna ，lnblna 1b4】pq1111x0,1xp1x 【证】令f(x)1xp1 (x f(x)xp1f(x0,x1,f(xp1)xp2,f(1)p1)x0,有f(x)f(1)0即1xp1 5设limf(x)1f(x)0f(x)x 【证1】由limf(x)1知f(0)0，f(0)1， f(x)f(0)f(0)xxf()

f()x (f(x)原式得证2】1f(0)0f(0)又f(x)0，则f(x)单调增，由日中值定理f(x)f(x)f(0)f(c)x （c介于0与x之间）由于f(x)单调增，则f(x)f(c)xf(0)x原题得证3】f(xx0F(x)f(xxF(x)f(x)1F(0f(01F(xf(x0，则F(x单调增，x0F(xx0FF(xf(x)0取得它在(,从而F(x)F(0)f(004】1f(0)0f(0)yf(x在(0,0)yf(x0yf(xyf(x在(0,0)yx的f(x)【例6】试证(xy)lnxyxlnxyln (x0,y0)2【证】xy

xyxlnxyln

(x0,y 即只要证函数f(x)xln (x0)的图形是凹的f(x)lnx

f(x)1x

(xf(x)xlnx(x0)的图形是凹的，原题得证7】pq0111x0y0xy1xp1yq 1）积分不等式性质；2）变量代换；3）积分中值定理； (bf(x)g(x)dx)2bf2(x)dxbg2(x)dx；

(11

1x

2【例1】设M 2

dx,N

dx,K 2 cosx)dx22 MNK MKN22 KMN KNM(1 12x M 2

1

dx 2

1

dx 22

1

0由不等式ex1x(x0)N 2

1x

dx

21dx2K 2

cosx)dx

21dx KMN故应选2】f(xf(1)f(1)1,f(0)1,f(x0,则（ （A）1f(x)dx （B）1f(x)dx （C）1f(x)dx0f （D）1f(x)dx0f3f(x[0,1] 求证:0f(x)dxa0f (0a 1】0f(x)dxa0f(x)dxaaf 即(1a)0f(x)dxaafa(1a)0f(x)dxa(1a)f(c11aaf(x)dxa(1a)f(c21f(xf(c1f(c2

0c1aac2a则(1aa

f(x)dxaaf111【证2 0f(x)dx=a0f （令xat1a0ff(xaxxf(axf 从而有a0f(ax)dxa0f 0f(x)dxa0f 【证3】令F(u)0f(x)dxu0f (0u F(u)

f(u)0f1f(u)f (0c1F(c0当0uc时，F(u0,当cu1时，F(u0F(u在区间[0,1]上的最小值必在u0或u1F(0)F(10,则当0u1F(u)0即 0f(x)dxa0f4】f(xg(x在区间[a,bf(x0g(x)x（I）0ag(t)dt(x xx （II）aag(t)dtf(x)dx bf 0ag(t)dta1dt(x x 令F(u) uf(x)g(x)dx aag(t)dtf 只要证明F(u)f(u)g(u)faug(t)dt g(u)f(u)faug(t)dt 由（I）的结论0ag(t)dtxaaaag(t)dtxuaaag(t)dtuf

f(u

fau(t)dtF(u0F(bg aag(t)dtf(x)dx bf 5f(x在[0,1f(0)0求证:1f2x)dx11f2 2【证】Qfxxf0f2(x)

xxf(t)dt xx

xf2 1 1 xxf2(t)dtx1f2 1f2(x)dx1xdx1f2(t)dt11f2(t)dt0

2 aa设a1,n1,(n1)2

analn

anf(x0f(00x10x20f(x1x2ln 设x0,x1,试 x1 设01，试证：xx1 (x0)

f(x1)f(x2)设1

dx,

dx,

tan（A）23（