2.2 几个著名不等式 2.2.1 著名不等式
柯西不等式 对于任意两组实数
a1,a2,?,an和b1,b2,?,bn有
2222(a1b1?a2b2???anbn)2?(a12?a2???an)(b12?b2???bn)
上述不等式只有当aa1a2????n时，等号才能成立． b1b2bn证明 因为对任意x，有
(a1x?b1)2?(a2x?b2)2???(anx?bn)2?0
222将上式展开得(a1?a2???an)x2?2(a1b1?a2b2???anbn)x?
22(b12?b2???bn)?0
上述二次三项式对任意x均大于等于0，故其判别式不能大于0，所以
2222(a1b1?a2b2???anbn)2?(a12?a2???an)(b12?b2???bn)
当判别式等于0时，上述方程有重根，设重根为x=k，则
(a1k?b1)2?(a2k?b2)2???(ank?bn)2?0
这时b1?ka1,b2?ka2,?,bn?kan 所以上述不等式只有当
bb1b2????n?k a1a2an时等号才能成立。
如令b1?b2???bn?1，则得
22(a1?a2???an)2?n(a12?a2???an) 22a12?a2???an(a1?a2???an)?n? n2222a1?a2???an2a12?a2???an()? nn22a1?a2???ana12?a2???an ?nn柯西不等式在高等代数中的意义是：两个向量的数积不大于两个向量长度的乘积．若
??(a1,a2,?,an),??(b1,b2,?,bn)
则
n
n(?,?)??? 2其中(?,?)??ab,???aiii?1i?1i,???bi?1n2i
例１ 若a1,a2,?,an都是正数，求证
?111?2? (a1?a2???an)??????n?aa?an?2?1证明 构造两个实数列
a1,a2,?,an;则由柯西不等式得
111 ,,?,a1a2an[(a1)2?(a2)2???(an)2]
222???1????1?1??? ????????????a??a????a1?2?????n??????a1?1?a2?1???an?1??n2 ?a1a2an???即
2?111?2? (a1?a2???an)??????n?aa?an?2?1?n??n2??n2???aibi????ai???bi? ?i?1??i?1??i?1?2*赫勒德尔不等式 由柯西不等式
nn?n?2可得??(aibi)(cidi)???(aibi)?(cidi)2
i?1i?1?i?1?nnn2但
(?aibi)?(?ai)(?bi)
22244i?1i?1i?1(?cidi)?(?ci)(?di)
22244i?1i?1i?1nnn所以有
(?aibicidi)?(?ai)(?bi)(?ci)(?di)
44444i?1i?1i?1i?1i?1nnnnn同理有(ab?f)????????iiii?1共8个实数列n8?(?ai)(?bi)?(?fi)
i?1i?1i?1???????????888共8个和式乘积nnn一般地有(aibi?li)????????i?1共2m个实数列n2m?(?ai)(?bi)?(?li) i?1i?1i?1?????????????共2m个和的积mn2mn2mn2m现在证明上述不等式对任意不等于2m的正整数k?2也成立（假定所有数列均为正数列）．
设
a1b1?g1设
a2?an?b2?bn???共k个实数列 ??g2?gn??k?2mk?Ai?ai,Ai?2mk?Bi共k个?Bi?bi???m2k??Gi?giGi?Hi2?aibi?gi??Hi?L(2m?k)?i????W?Hi?i
m?a2,i?1,2,?,nkm?b2,i?1,2,?,n??km
?g2,i?1,2,?,nm1m1m1m再令
Hi?a2b2?g2,i?1,2,?,n
则有
(?i?1nAiBi?GiHiLi?Wi)2?
??????????共k个共(2m?k)个m(?A)(?B)?(?G)(?Hi2)2i?1i?1i?1i?1n2min2min2minmm?k
kkmkm1m1m1m但
AiBi?GiHiLi?Wi?a2b2?g2(a2b2?g2)2m??k
m
?aibi?gi
所以
????aibi?gi??i?1?n2m???????????aik???bik????gik???aibi?gi??i?1??i?1??i?1??i?1?nnnn2m?k
所以
?n??nk??nk??nk???aibi?gi????ai???bi????gi? ?i?1??i?1??i?1??i?1?k即该不等式对任意不等于2m的整数k也成立．
上述不等式的证明有些麻烦，不好记，现用反归纳法给出一个简洁的证明． 由证明知，不等式
(?aibi?gi)?????i?1共2m个n2m?n2m??n2m??n2m????ai???bi????gi? ?i?1??i?1??i?1?对无穷多个自然数k=2m成立．
现在假设不等式对m=k成立．
[?(ab?g)(ab?g)]k（是k个数列）≤
???????i?1k?1个数列?1k?1k?111nn?nkk??n1?kkkkkkkkk??(ai)?(bi)??(gi)???(aibi?gi)?
i?1i?1?i?1??i?1?nk?1k?1kkiik?1ki11kkii1ki但是
?1k?1k?1111?nkk?kkkkk左边???aibi?gi(aibi?gi)?
?i?1?k
?n????aibi?gi? ?i?1??nk?1??nk?1??nk?1??n????ai???bi????gi???aibi?gi? ?i?1??i?1??i?1??i?1?k?1k
所以
?n???aibi?gi??i?1??nk?1??nk?1??nk?1????ai???bi????gi? ?i?1??i?1??i?1?即不等式对m=k-1也成立。由反归纳法知，不等式对任意整数k均成立．
例２ 设非负实数x1,x2,?,xn满足
x1?x2???xn?求证(1?x1)(1?x2)?(1?xn)?1 21． 2证明 当n=1时，结论显然正确．
假设命题在n=k时正确，非负实数x1,x2,?,xk满足
x1?x2???xk?1 2则(1?x1)(1?x2)?(1?xk)?1成立． 2现设x1,x2,?,xk,xk?1为k+1个非负实数，满足
要证
1 21(1?x1)?(1?x2)???(1?xk)?(1?xk?1)?
2x1?x2???xk?xk?1?令x??xk?xk?1，则由归纳假设
1 21(1?x1)(1?x2)?(1?x?)?
2x1?x2???xk?1?x??但是，因为xk?xk?1?0，所以
1?x??1?xk?xk?1?1?xk?xk?1?xk?xk?1
?(1?xk)(1?xk?1)
1 2所以
(1?x1)(1?x2)?(1?xk)(1?xk?1)?(1?x1)(1?x2)?(1?x?)?证毕 如果令ai??,bi??i?gi??1ki1k1ki．
这里?i?i??i均为正实数，则得
????(??)(??)???????iiii?1共k个数列i?1i?1n1ki1k1kin1kn1k?(??i)
i?1n1k现在证明下面不等式
??i?1n?i?i??i?(??i)(??i)?(??i)?
????i?1i?1i?1nnn其中?,?,?,?均为正有理数，且
????????证明
1ki1ki?1??1????1k1ki1ki1k?k?1 k1ki1ki1ki1ki??i?1n?i?i??i????????i??i??????
??i?1n?????????????????共?1个共?1个共?1个1k