文档介绍：第十九讲3学时3、在不等式两端取变限积分证明新的不等式例15证明:当x0时,3Xx—sinx63Xx—65X120证已知cosx1, (X0,只有2n时,等号成立)在此式两端同时取[0,x]上的积分,Xcostdt0X0dt,sinx(x0)再次取[0,x]上的积分,得2Xcosx—第三次取[0,x]上的积分,得sinx所以3XX一6sinx上式再在[0,x]上的积分,得4X24cosxcosx2x_24x_24再在[0,x]上的积分,得sinx3XX+(x)是[a,b]:[a,b],为 X2,有几何解释:X1X221X2 X1X2xf(t)dtx1f(X1)f(X2)X1(X2%), (0,1),则X2X2 X-I X11f(t)dt0f(x1 (X2xj)d(1)同理,令t X2 (X2Xi), (0,1),则1 X2 1Lxif(t)dt 0f(X2 (X2Sd从而X2,f(t)dtf(X1(X2X1))f(X2 (X2X1))d注意到X1 (X2X1)与X2 (X2xJ关于中点勺一X2对称,f(x)又2为凸函数,所以1f(X1 (X2X1)) f(X22(X2X1))Xi X2另一方面,由(1 X2Xf(t)dtX2 X1X11)式及f(X)的凸性X-I X2).tX1(X2X1) 1f(t)dt 0f(X2(1 )xjdf(X2)(1)f(X1)(X2)2f(X1)f(X1)f(X2)2例17设函数g(X)在[a,b]: x(a,b)X函数f(x)g(t)(x)在[a,b]上递增,X1,X2,X3 (a,b),X1X2 X3f(X2)f(X1) 1X2X2X1X2X1g(t)dtX1cg(t)dt1 X2 1 X3xg(t)dt g(X2) xg(t)dtX?X1X1 X3X2X3 X2cg(t)dtcg(t)dt f(X3)f(X2)X3X2 X3X2所以,f(x)(x),p(x)在[a,b]上连续,p(x)0,bpgdxa且mf(x)M, (x)在[m,M]上有定义,并且有二阶导数,(x) :p(x)f(x)dxbaPgf(x)dxp(x)dxp(x)dx令n,得baPfiPi(fi)baPiPibp(x)f(x)dxabbp(x)f(x)dxabp(x)dxp(x)dx证II(利用Taylor公式)记x0p(x)f(x)dxxi a_(bna),PP(xJ,fif(Xi),(i1,2,L,n)Q(x)0,(x),取Pii n,(j1,2丄n,n),i 1Pji1j1nni1ifiii1(fi)P1f1p2f2L PnfnP1(fjP2⑹L Pn(fn)(利用积分和)将区间[a,b]n等分,记pi P2 L Pn Pi P2 LPnp(x)dx则 (y) (x0) (x0)(yx0)2()(yx0)()0,(y) (xo)(x°)(yX。).在上式中,令yf(x),然后两边乘以P(X) 得b ,得p(x)dxap(x)f(x)bap(x)dx(Xo)bp(x)ap(x)dxgpmf(x)xobaP(x)dx在[a,b]上取积分bp(x)abap(x)dx(X。):bp(x)dxap(x)dxbaP(x)其中aP(x)(X。)bp(x)f(x)x。abp(x)dxadxf(x)dxbp(x)dxaf(x)x0dx(x。)baP(x)b(Xo)ap(x)f(x)X0dxbp(x)dxaf(x)bap(x)f(x)dx—r dxp(x)dxa1baP(x)dxbaP(x)dxbap(x)f(x)dxab baP(x)dxap(x)f(x)dxa ap(x)f(x)dx(Xo)p(x)f(x)dxp(x)dxp(x)dx§、(i1,2丄,n)为任意实数,则n na" i1(Cauchy不等式)(判别式法)n20aixbi1n2 2cai x2i1nahi1nb2i1上式是关于x的二次二项式,保持非负,故判别式nabi1n2aii1nb2 (配方法)nQa2i1nb2i1nahi12ainb2j1naibii1najbjj1n22aibjj1nabajbjj11nn〔,©ajbi,(i1,2,L,n).证III(利用二次型)n2oa/byi1n2aii1n2abii1xynb2i1即关于x,y的二次型,非负定,因此n2aii1n2aii1naibii1nabii1nb2i1b2i