题13 设x1，x2，x3，y1，y2，y3是实数，且满足x1?x2?x3?1，证明不等式
2222(x1y1?x2y2?x3y3?1)2?(x12?x2?x3?1)(y12?y2?y3?1).
222 （第十届高二第二试第22题）
证法1 当x1?x2?x3?1时，原不等式显然成立.
222当x1?x2?x3?1时，可设f?t??x1?x2?x3?1t?2?x1y1?x2y2?x3y3?1?t
2222222??22??y12?y2?y3?1?.易知右边??x1t?y1???x2t?y2???x3t?y3???t?1?.
2222?f?1???x1?y1???x2?y2???x3?y3??0.?f?t?是开口向下的抛物线，
2222222??t?4?x1y1?x2y2?x3y3?1??4?x12?x2?x3?1??y12?y2?y3?1??0即
22222(x1y1?x2y2?x3y3?1)2?(x12?x2?x3?1)(y12?y2?y3?1).
综上，x1?x2?x3?1时，
2222(x1y1?x2y2?x3y3?1)2?(x12?x2?x3?1)(y12?y2?y3?1).
222证法2，?xi,yi?R?i?1,2,3?，x1?x2?x3?1，?当y1?y2?y3?1时，
?2222222222(x12?x2?x3?1)(y12?y2?y3?1)?0，又(x1y1?x2y2?x3y3?1)2?0，?求证的不等式
成立.当y1?y2?y3?1时，(x1?x2?x3?1)(y1?y2?y3?1)?
222222222?1?x212222?x2?x31?y12?y2?y3????2222?1?x12?x2??x3?1?y12?y2?y3????2??2???2??x12?y12??x22?y22??x32?y32??21????1?xy?xy?xy???????????112233222????????(x1y1?x2y2?x3y3?1)2.综上，在题设条件下，总有
2222(x1y1?x2y2?x3y3?1)2?(x12?x2?x3?1)(y12?y2?y3?1).
证法3 设a?x1?x2?x3?1，b??2(x1y1?x2y2?x3y3?1)，c?y1?y2?y3?1，则
2222由x1?x2?x3?1知a?0，从而a?b?c?x1?x2?x3?1?2?x1y1?x2y2?x3y3?1??y1
222222222 6
22?y2?y3?1??x1?y1???x2?y2???x3?y3??0.
222?b2?4ac??2a?b???4a2?4ab?4ac??4a?a?b?c??0,b2?4ac??2a?b??0，
22?2222?x3?1)(y12?y2?y3?1). ?b2?4ac?0，即(x1y1?x2y2?x3y3?1)2?(x12?x2??证法4 设a??x1,x2,x3?，b??y1,y2,y3?，则 ??a?b??x1,x2,x3??y1,y2,y3??x1y1?x2y2?x3y3，又 ????a?b?a?b?cos??2222x12?x2?x3?y12?y2?y3?cos?.
2222?x1y1?x2y2?x3y3?1?1?x12?x2?x3?y12?y2?y3?cos??222222221?x12?x2?x3?y12?y2?y3?cos??1?x12?x2?x3?y12?y2?y3?0 22222?(x1y1?x2y2?x3y3?1)2?(1?x12?x2?x3y12?y2?y3)
2222?(x12?x2?x3?1)(y12?y2?y3?1).
B??y1,y2,y3?，O?0,0,0?为坐标原点，证法5 记A??x1,x2,x3?，则由AB?OA?OB,
得?x1?y1???x2?y2???x3?y3?222?2222，整理得 x12?x2?x3?y12?y2?y322221??x1y1?x2y2?x3y3??1?x12?x2?x3?y12?y2?y3?0，
22?xy1?x2y2?x3y3?1?1?x12?x2?x3222y12?y2?y3?0，
2223?(x1y1?x2y2?x3y3?1)?1?x?x?x?212223y?y?y21??(x?x?x?1)(y?y?y?1).
2212223212223评析 这是一个条件不等式的证明问题.由求证式是b?ac的形式自然联想到二次函数的判别式，构造一个什么样的二次函数是关键.当然是构造
2222f?t??x12?x2?x3?1t2?2?x1y1?x2y2?x3y3?1?t?y12?y2?y3?1，但只有当 2222x12?x2?x3?1?0时，f?t?才是二次函数，故证法1又分x12?x2?x3?1?0与
22x12?x2?x3?1?0两类情形分别证明.很显然，等价转化思想、分类讨论思想是证法1的精髓.
2????证法2直接运用基本不等式证明.证法3通过换元后证明b?4ac?0（即求证式），技巧性很强，
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2一般不易想到，读者可细心体会其思路是如何形成的.证法4由求证式中的
22222x12?x2?x3,y1?y2?y3及x1y1?x2y2?x3y3联想到空间向量的模及数量积，因而构造向量
解决问题.证法5则从几何角度出发，利用AB?OA?OB使问题轻松得证.五种证法，从多角度展示了本压轴题的丰富内涵.
拓展 本题可作如下推广：
推广 1 若xi,yi?R?i?1,2,?,n?,2?xi?1n2i?1，则
?n??n2??n2???xiyi?1????xi?1???yi?1?. ?i?1??i?1??i?1?推广 2 若xi,yi?R?i?1,2,?,n?,m?0，
?xi?1n2i?m，则
?n??n2??n2???xiyi?m????xi?m???yi?m?. ?i?1??i?1??i?1?两个推广的证明留给读者.
2x2y2z2???2， 题14 已知x、y、z?0，并且
1?x21?y21?z2x2y2z2???2. 求证：2221?x1?y1?z （第一届备选题）
证法1 令x?tan?,y?tan?,z?tan?，且?,?,?为锐角，则题设可化为
sin2??sin2??sin2??2，即co2?s?2c?o?s2?c?os柯西1不等式知.由
2?2?1??sin2??sin2??sin2???cos2??cos2??cos2??
??sin?cos??sin?cos??sin?cos??2?1????sin2??sin2??sin2???. ?2?2?1?sin2??sin2??sin2???2.由万能公式得 2tan?tan?tan?xyz???2???2. ,即
1?tan2?1?tan2?1?tan2?1?x21?y21?z2
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证法2 构造二次函数
?1x??1y?f?t???t??t???2221?x1?x1?y2???1?y22??1z????t??2??1?z21?z??2?1?x11?2yz?222??xyz????t?2??t???222222??? . ?222??1?x1?y1?z??1?x1?y1?z?1?x1?y1?z???f?t??0，当且仅当x?y?，即?,z取t?x?y?z时取等号，??0?xyz?2?111??x2y24??1?x2?1?y2?1?z??2?4??1?x?z2?12?y?1?2z????1?2x?1?y2?1?z2???0?11?x2?1?x21y21z21?x2,1?y2?1?1?y2,1?z2?1?1?z2, 又x2y2z211?x2?1?y2?1?z2?2,?1?x2?11?y2?11?z2?1, 2?4??xyz??1?x2?1?y2?1?z2???4?1?2?0， 故
x1?x2?y1?y2?z1?z2?2.（当且仅当x?y?z?2时取等号） 证法3 x21?x2?y21?y2?z21?z2?2， 即
???1?1?1?x2??????1?1?1?y2??????1?1?1?z2???2，即11?x2?11?y?121?z?1,2于??1?1?x2?1?1y??1?x2y2z?2?xy?22?1?, z???z2?1?x?2?1?y2?????1?z2??1?x2?1??y21z2即
x1?x2?yz1?y2?1?z2?2. 证法4 令x2y2z21?x2?X,1?y2?Y,1?z2?Z,则X?Y?Z?2，且 9
，
是2?xyz??XYZ?XYZ??????? ，所以?x2?,y2?,z2?222?1?X1?Y1?Z?1?x1?y1?z??xyz?22??222?X?XYZ?YZ?222? ?3?X?Y?Z?X?Y?Z?3?2?2?2??3??????????XYZyz??x???1?X1?Y1?Z?222xyz12??1???3?2??X?Y?Z???3?2??22??2.所以???2. 222331?x1?y1?z????x22ay22bz22c?,?,?, 证法5 设2221?xa?b?c1?ya?b?c1?za?b?c则x?22a2b2c,y2?,z2?,
b?c?aa?c?ba?b?cx21y21z21????? 左边=
1?x2x1?y2y1?z2z?1b?c?ac?a?ba?b?c???2b??2c??2a???a?b?c?2a2b2c??????2a?b?c?a?b?c?a??b?a?c?b??c?b?a?c??
2?3?a?b?c?a??b?a?c?b??c?b?a?c?a?b?c6?2?ab?bc?ca???a2?b2?c2?a?b?c62122??a?b?c???a?b?c??2.a?b?c33x22证法6 ???2221?x1?x2x2?1?x2?2x?22?; 21?xy22yz22z??22?;??22?. 同理?1?y21?y21?y21?z21?z21?z2?1x2y2z211????2??三式相加得 ?222222?1?x1?y1?z?1?x1?y1?z? 10