第十三章函數(shù)方程與不等式證明

第十三章1a a1

an1an.(a>1, 一.證明不等式(n 證明:令f(x)ax,在 ,1上使用拉格朗日定n f(

)f( )f'()(1 1n n 1n an1aln

n(n ana

n(n a ana(n

a

.(a>1,n二.a0,b0,0p1,(ab)papb證明:fx)xbpxpf(0)0.x0時(shí),0pf'(x)p(xb)p1pxp1x0時(shí)f(x)單減,f(a)f(0)0.(ab)papbp即得(abpapb三.f(x)在[0,1]上有連續(xù)導(dǎo)數(shù),滿足0fx1且f(0)0. 0f(x)dx0f( :

f(t)dt

f(t)dt,F(0)=0.因?yàn)?fx1且f(0)0x0f(x)F'(x)2f(x)xf(t)dtf3(0 f(x)2f(t)dtf(x) 令x)2xf(t)dtf2x,顯然(0)=0'(x)2f(x)2f(x)f'(x)2f(x)(1f'(x))x>0時(shí),(x)>0.由(1)Fx0(x>0).x>0F(x)F(0)=0.F(1)F(0)0. 0f(x)dx0f(四.求證|a|p|b|p21p(|a||b|)p0p求證:0x1,0p1時(shí),21pxp(1x)pFxxp1x122F'(x)pxp1p(1x)p1.F'(x)0得x .F(1)21p，F(xiàn)(1)F(0)11222F121p為最大值，F(xiàn)(1)F(0)1.0x1,0p1時(shí),221pxp(1x)p2令x |a|a||b

,則1x |b|a||b

.代入上述結(jié)論,21p

|a(|a||b|)

|b(|a||b|)p (|a||b|)p|a|p|b|p21p(|a||b|)p,(0<p<五.求證:xyz6,x2y2z212x0,y0,z2(x2y2z2)2xy2yzx2y2z2(xyz)22xy2yz362(x2y2z2所以3(x2y2z2)36 x2y2z2s(x,y,z)x2y2解以下條件極值問(wèn)題條件：xyzF(xyz,x2y2z2－(xyFx'2x0,Fy'2y0,Fz'2z解得xyz2.只有一個(gè)駐點(diǎn),xyz212.x2y2z212,(x0,y0,z六.證明1f(x)在[a,b]fx0，則(ba)f(a)bfx)dx(ba)f(af(b) 2f(x)在[a,b]fx0(ba)f(b)bf(x)dx(ba)f(a)f 證明：11:f(x)是增加的,所以對(duì)于[a,b]x,f(xf(a),baf(x)dxf(a)(b f(a)f(令F(x)af(t)dt(x F'(x)f(x)f(a)f(x)(xa)f'(x)f(x)f(a)(xa)f'( f'()(xa)f'(x)(xa (a xaffx0(fx)02F(x)單增.F(a)0,F(b)F(a)0.bf(x)dx(ba)f(a)f 2:f(x)t, f(t)f(x)f'(x)(tx)f''()(t所 f(a)f(x)f'(x)(ax)f''(1)(af(b)f(x)f'(x)(bx)f''(2)(bf(a)f(b)2f(x)f'(x)(ab)2xf'(x)f''(1)(ax)2

f''(2)(b (f(b)f(a))(ba)2af(x)dx(ab)af'( bb

xf'(x)dx

bf''(1

(a

f''(2

2 (b (bb

(f(b)f(a))(ba)2af(x)dx(ab)(f(b)f(a))2xf(

2af(abbbb

f(x)dxaf(b)af(a)bf(b)bf(a)2bf(b)2aff(x)dx(ba)(f(a)fb于 2(ba)(f(b)b

f(a))b

f(即 (f(b)f(a))即

f( 注:12,f(x)單增的條件x2x2 x2x2 nn七.證明: 2 n x1x2Lxnn 證明1方法一:先證aba2k

kk

k

kn由(akxn

2 k)2akx2akbkxb2kk

k k k 得 aba2k

kk

k

k上述不等式中令ax，b1, xkk

x2n

kk kxxL x2x2L 2

n nn方法二:令f(x)x2

Lpn因 f''(x)2所 f(p1x1Lpnxn)p1f(x1)Lpnf(xnxL x2L (

n) x2x2 x2x2 nn 2 n1 取f(x)=lnx,f''(x) 0.令p1=p2=?=pn=所以lnx1Lxn1lnxL1lnx 立即得到x1x2Lxn n八.fxc[ab,f(a)f(b)0,求證bf(x)dx(ba)3max|f''(x)

a證明:1:1bfxxaxb)dx1bxaxb)df2 2b=1(xa)(xb)f'(x)

1

f'(x)(2xabbb=1(2xab)df(x)bbb

f(x)(2xa 1

f(2

2b=af(b1 1所以|a

f(x)dx| 2

f''(x)(xa)(xb)dx

|f''(x)(xa)(xb)|21max|f''(x)|b(xa)(bx)dx=(ba)3max|f''(x)2a f(t)f(x)f'(x)(tx)

a2(t2 2所以0=f(a)f(x)f'(x)(ax) (af(x)f'(x)(xa)f''()(abf(x)dxbf'(x)(xa)dxbf''()(xa)2 =f(x)(xa)bf(x)dx1f''()(xa)2 2a所以bfx)dx1bfxa)2 4于是|bfx)dx|1b|f|xa)2dx1max|fx|bxa)2 4(b= max|f''(x) a

4a 九.f(x)在[a,b]上二階可導(dǎo),x[a,b]fxfx0. b|f(x)|dx1|f(a)f(b)ba 證明:f(x)在[a,b]上二階可導(dǎo),f(x)連續(xù).x[a,b]fxfx0x[a,b]時(shí)f(x)0.分二種情形x[a,b]時(shí)f(x0.fxfx0fx0.bf(x)dx(ba)f(a)fa f2(a)f所 af(x)dx(b , b|f(x)|dx1|f(a)f(b)ba x[a,b]時(shí)f(x0.fxfx0fx0.2bfx)dx(ba)f(af(b),2ab |f(x)|dxb

1|f(a)f(b)ba 注:原題為:f(x)在[a,b]上二階可導(dǎo),x[a,b]fx0. b|f(x)|dx1|f(a)f(b)ba 該題是錯(cuò)誤的:f(x)x212x f''(x)20 b2,a2.計(jì)算可 b|f(x)|dx2[1(x21)dx2(x21)dx]b |f(b)f(a) |f(2)

f(2)| baa|fx|dx2|f(af(b希望讀者自己找出書(shū)中所給出的證明中的錯(cuò)誤十.設(shè)x>0,證 ln11 x2x22x

x證明:令F(x) ln11.于是limF(x)02xx>0時(shí),Fx)

4 x4

1 (2x

x x

(2x x(x(x1(x12

x(xx0F(x)單增.x0F(x0. ln112x x

1x .于是limx)01x x(x當(dāng)x>0時(shí),'(x) x(x2x

x(x

2x(x x(x1),所以'(x)0,所以當(dāng)x>0時(shí)(x)0.2ln1

1 x2x2 x十一.fx在[0,2]上連續(xù),fx0,n(正整數(shù)) f(x)sin

2[f(2)fn1證明:0f(x)sinnxdx=n f(x)dcos 1 n= (f(2)f(0)) f'(x)cos nn f(2)f 1 n所

f(x)sinnxdx

n f'(x)dx [f(2)f十二.設(shè)在[ab]fx)0,ax1x2<b,01,試證f(x1)(1)f(x2)f[x1(1)x2證明:f(x1(1)x2fx2(1x2x1f1f(x2)f(x1(1)x2)(x2x1)f'(2

f0,f(x1)(1)f(x2)f[x1(1)x2十三.設(shè)函數(shù)P(x,y),Q(x,y在光滑曲線C上可積,L為曲線C的弧長(zhǎng),而P2P2Q(x,y

.

PdxQdyLMCPdxQdyPdxQdydl{PQ{coscos C P2P2Q cos2cos2C

d