文档介绍：利用柯西不等式证明不等式2,利用柯西不等式证明不等式aaa,,,,,bbb,,,,柯西(Cauchy)不等式:设和是给定的实数,则12n12n2222222。等号成立的充要条件是:存在(不,,,()()()abababaaabbb,,,,,,,,,,,,,,,,,,,nnnn11221212,,ab,全为零),使i=1,2,???nii证明:nnnn2222设fxaxabxbaxb()2()0,,,,,,,,,,iiiiii,,,,1111iiiinnn222则:,包括全为的情况。化简即得:(),,,0(0)()()aabab,,,iiiii,,,111iii3)柯西不等式的几个推论(abRin,,1,2,,,,?若,ii2n,,a2,i222,,2nxx,x,xxx,,a,i1,,312312ibain,,,,1,2,,则,当且仅当时取等号。特别地,,,,,ii,nyyyy,y,yb,i1123123ib,i,i12n,,a,i,,na,i1,,ibbb,,,?,当且仅当时等号成立,12n,nb,i1iab,ii,i12nn,,2,?若,则naa,,aRin,,,1,2,3,,,,iii,,,,,,ii11aaa,,,当且仅当时等号成立12nnn,,1,,2,aaa,,,?若,则an,当且仅当时,aRin,,,1,2,3,,,,,,12nii,,a,,11ii,,i,,222abcabc,,,,,例1:若都是正数,证明:abc,,bccaab,,,2222,,abc2证明:由,,,,,,,,,,bccaababc,,,,,,,,,,,,,,bccaab,,,,,222abcabc,,,,,aab,,,2xxxn12xxx,,,例2:设为正实数,证明:,,,,n12n222222111,,,,,,,xxxxxx11212n21,,,,xxx,,n12222证明:,故有1,,,,,xxxn12n1n,22221,,,,xxx1,,,,xxx,,n12n12n,11,,,,,,,,xxxxxx,,,,12112nn,,,n11,,,,xxxxxxx11,,,,,,,,nnn12112,,,aaa,,,,,例3:是个不同的自然数,求证:n12naa11n2,1,,,,,,a,,,,,,12222nn22,,aa11111,,n21,,,,,,,,,,,,,,,,a证明:?,,1,,22nnaaa,,,,12n,,,,aa111,,n2,,,,,,,,,,,,,a,,1,,222naaa,,12n,,aa11,,,,n2,,,,,,,,,,,,,a11,,,,2222nn,,,,aa11n2,?1,,,,,,a,,,,,,12222nn说明:本题通过变形积极创造了柯西不等式中“积和”与“平方和”的形式。*例4:已知且,求证:nN,n,24111112,,,,,,,,,,,172342122nn,11111证明:?1,,,,,,,,,234212nn,111111,,,,,,,,,,,,,,,,,,,12,,,,232242nn,,,,11111,,,,,=1,,,,,,,1,,,,,,,,,,232n2n,,,,111,,,,,,,nnn,,12241112?原不等式,,,,,,,,,71222nnn,,由柯西不等式有:111,,2,,,,,,,,,,,,,,,nnnn122,,,,,,,,,,nnn,,122,,2n111?,,,,,,,nnn,,,,,,,,122nnn,,122,,,,24,,173,n又由柯西不等式有:2,,111111,,222,,,,,,,,,,,,,,,,,,,111,,,,,,222nnn,,122,,nnn,,122,,,,,,,,,,,,111,,,,,,,n,,nnnnnn,,,,112212,,,,,,,,,,111,,,,,n,,nn22,,2111?,,,,,,,2nnn,,122故原不等式成立。说明:本题是一道利用柯西不等式对连续自然数倒数和估值的典型例题。本题有一定的难度,第一步代数变形是基本功,将式子化为若干项之和,、三步对柯西不等式的两种不同的运用堪称范例,值得回味。1119例5:已知为正实数,且,,,a,,,1,,,222abc,,,1114222abc3,,,证法1原不等式等价于222abc,,,1114由柯西不等式,可得2222abcabc(),,,,,222222abcabc,,,,,,,,1111112()abc,,,a,,,,,3()2()abc,,,2()()a,,,,,2()3abc,,.,,1422()()abcabc,,,,,3222abc3证法2,,,222abc,,,1114222abc3,,,.,()()()()()()4abacb