Chebyshev不等式的应用_蔡玉书

1、Chebyshev 不等式的应用蔡玉书(江苏省苏州市第一中学,215006本文先给出著名的车比雪夫(Cheby shev 不等式,然后用它证明一些不等式竞赛试题.C hebyshev 不等式设a 1a 2a n ,b 1b 2b n ,则n k =1a k n k =1b k n nk =1a kb k(1设a 1a 2a n ,b 1b 2b n ,则nk =1a k nk =1b k n nk =1a kb k(2证明n nk =1a k b k -nk =1a k nk =1b k =nk =1nj =1(a k b k -a k b j =nj =1nk =1(a j b j -a j

2、 b k =12n k =1nj =1(a k b k +a j b j -a k b j -a j b k =12n k =1n j =1(a k -a j (b k -b j 0.即得(1.(2留给读者自己完成.下面给出定理的应用.例1(2008年塞尔维亚数学奥林匹克试题已知a ,b ,c 是正数,且a +b +c =1,证明:1bc +a +1a +1ca +b +1b +1ab +c +1c2731.证明1bc +a +1a+1ca +b +1b+1ab +c +1c27319a 2+9abc +9-31a a 2+abc +1+9b 2+9abc +9-31b b 2+abc +1+

3、9c 2+9abc +9-31c c 2+abc +10不妨设a b c ,显然9(a +b <31,所以容易证明9a 2+9abc +9-31a 9b 2+9abc +9-31b 9c 2+9abc +9-31ca 2+abc +1b 2+abc +1c 2+abc +1.即1a 2+abc +11b 2+abc +11c 2+abc +1由Chebyshev 不等式有3(9a 2+9abc +9-31a a 2+abc +1+9b 2+9abc +9-31bb 2+abc +1+9c 2+9abc +9-31cc 2+abc +1(9a 2+9abc +9-31a +(9b 2+9a

4、bc +9-31b +(9c 2+9abc +9-31c (1a 2+abc +1+1b 2+abc +1+1c 2+abc +1于是只要证明(9a 2+9abc +9-31a +(9b 2+9abc +9-31b +(9c 2+9abc +9-31c 0.它等价于9(a 2+b 2+c 2+27abc +27-31(a +b +c 0.因为a +b +c =1,只要证明9(a 2+b 2+c 2+27abc -40因为a +b +c =1,不等式等价于9(a 2+b 2+c 2(a +b +c +27abc-4(a +b +c 305(a 3+b 3+c 3-3(a 2b +ab 2+b 2

5、c +bc 2+ca 2+ac 2+3abc 0由舒尔(Schur 不等式a 3+b 3+c 3+3abca 2b +ab 2+b 2c +bc 2+c a 2+ac 2由均值不等式得a 3+b 3+c 33abc×3+×2得不等式,从而原不等式得证.例2(2006年中国女子数学奥林匹克试题60数学通讯2010年第7期(下半月·课外园地·设x i >0,i =1,2,n .k 1,求证:ni =111+x i ni =1x ini =1x k +1i 1+x i ni =11x k i.证明不妨设x 1x 2x n >0,则1x k 11x

6、k21x k n ,x k 11+x 1x k 21+x 2x k n1+x n (k 1.于是,根据Chebyshev 不等式,有ni =111+x i ni =1x i=(11+x 1+11+x 2+11+x n (x 1+x 2+x n =(1x k 1·x k 11+x 1+1x k 2·x k 21+x 2+1x k n ·x k n 1+x n ·(x 1+x 2+x n 1n (1x k 1+1x k 2+1x k n (x k 11+x 1+x k 21+x 2+x k n1+x n(x 1+x 2+x n (x 1·x k 1

7、1+x 1+x 2·x k 21+x 2+x n ·x k n1+x n ·(1x k 1+1x k 2+1x k n=(x k +111+x 1+x k +121+x 2+x k +1n 1+x n (1x k 1+1x k 2+1x k n=n i =1x k +1i 1+x i n i =11x k i.例3(2007年南斯拉夫数学奥林匹克试题已知x ,y ,z 是正数,且满足x +y +z =1,k 是正整数,证明:x k +2x k +1+y k +z k +y k +2y k +1+z k +x k +z k +2z k +1+x k +yk 17.证

8、明不妨设x y z ,则有x k y k z k ,由Chebyshev 不等式得3(x k +1+y k +1+z k +1(x +y +z (x k +y k +z k因为x y z ,则有x k +1+y k +z k y k +1+z k +x kz k +1+x k +y k事实上,x y z ,有x k -1y k -1z k -1,x (1-x -y (1-y =x (y +z -y (z +x =z (x -y 0,即x (1-x y (1-y ,于是x k (1-x y k (1-y ,从而x k +1+y k +z k y k +1+z k +x k ,同理y k +1+z

9、 k +x k z k +1+x k +y k .所以又有x k +1x k +1+y k +z k y k +1y k +1+z k +x k z k +1z k +1+x k +y k.所以,由Cheby shev 不等式得x k +2x k +1+y k +z k +y k +2y k +1+z k +x k +z k +2z k +1+x k +y k 13(x +y +z (x k +1x k +1+y k +z k +y k +1y k +1+z k +x k+z k +1z k +1+x k +y k=13(x k +1x k +1+y k +z k +y k +1y k +1+

10、z k +x k +z k +1z k +1+x k +y k=13(x k +1x k +1+y k +z k +y k +1y k +1+z k +x k+z k +1z k +1+x k +y k (x k +1+y k +z k +(y k +1+z k +x k +(z k +1+x k +y k ·1xk +1+yk +1+zk +1+2(x k +y k +z k (x k +1+y k +1+z k +1·1xk +1+y k +1+z k +1+2(x k +y k +z k=(x k +1+yk +1+zk +1·1xk +1+yk +1+zk

11、 +1+2(x +y +z (x k +y k +z k (x k +1+y k +1+z k +1·1xk +1+yk +1+zk +1+2×3(x k +1+y k +1+z k +1=17.最后一步用的是不等式.例4(2009年波兰数学奥林匹克试题已知a ,b ,c 是正实数,n 1是正整数,证明:a n +1b +c +b n +1c +a +cn +1a +b61·课外园地·数学通讯2010年第7期(下半月(a n b +c +b n c +a +c n a +b na n +b n +c n3.证明不妨设a b c >0,则因为ab +

12、c -b c +a =a (c +a -b (b +c (b +c (c +a =(a -b (a +b +c (b +c (c +a 0,b c +a -c a +b =(b -c (a +b +c (b +c (c +a 0,所以a b +c b c +a c a +b .当n =1时,由Chebyshev 不等式得3(a 2b +c +b 2c +a +c 2a +b (a b +c +bc +a +c a +b(a +b +c ,所以a 2b +c +b 2c +a +c 2a +b (a b +c +b c +a+c a +b (a +b +c 3,即n =1时,不等式成立.当n 2

13、时,由柯西不等式得(a n +1b +c +b n +1c +a +c n +1a +ba n -1(b +c +b n -1(c +a +c n -1(a +b (a n +b n +c n 2又由赫尔德(H o ··lder 不等式得(a n +1b +c +b n +1c +a +c n +1a +b n -1(a b +c +b c +a +c a +b (a nb +c +b nc +a +c na +bn 由,得(a n +1b +c +b n +1c +a +c n +1a +bn a n -1(b +c +b n -1(c +a +c n -1(a +b (

14、a b +c +b c +a +ca +b (a n b +c +b n c +a +c n a +b n (a n +b n +c n 2又因为a n -1(b +c -b n -1(c +a =c (a n -1-b n -1+ab (a n -2-b n -20,b n -1(c +a -c n -1(a +b =a (b n -1-c n -1+bc (b n -2-c n -20,所以,a n -1(b +c b n -1(c +a c n -1(a +b ,注意到a b +c b c +a ca +b,于是由Chebyshev 不等式得3(a n +b n +c n a n -1(

15、b +c +b n -1(c +a +c n -1(a +b (a b +c +b c +a +ca +b×得,3(a n +1b +c +b n +1c +a +c n +1a +b n (a nb +c +b nc +a +c na +bn (a n +b n +c n ,即a n +1b +c +b n +1c +a +c n +1a +b (a n b +c +b n c +a +c na +b na n +b b +c n3.例5(2008年土耳其集训队试题方程x 3-ax 2+bx -c =0有三个正数根(可以相等,求1+a +b +c 3+2a +b -cb的最小值.证

16、明设方程x 3-ax 2+bx -c =0的三个正数根(可以相等分别为p ,q ,r ,则由韦达定理得a =p +q +r ,b =pq +qr +rp ,c =pqr ,所以1+a +b +c3+2a +b=(1+p (1+q (1+r (1+p (1+q +(1+q (1+r +(1+r (1+p ,c b =pqr pq +qr +rp .记f (x ,y ,z =xyz xy +yz +zx .问题就是求f (p +1,q +1,r +1-f (p ,q ,r 的最小值.不妨设p q r >0,由Chebyshev 不等式得3(1p (p +1+1q (q +1+1r (r +1(1p +1q +1r (1