即待证 (1-yy) (1-xx)2 (1-xx) (2) 在证明(2)成立前,先证明以下的一些结果 引理1命μ=1+2ay+aa'yy,则 (3) 证由(3.7)已知 +(1=2a)y+y2a) 十2a 即待证 a 1 a ykt yy ak 左边等于 (δk+2y;aA)(1+2ay+aa’y) k ak)o a y -1z {I2a(1+2ay+ayy")十2(δ;+2y;ak) (a1+aa'y)一2(6+2yak)(a1+aa’y;) 2(yk yy'ak)a’](1+2ay+ 2n[(+2y;ak)(1+2ay2+ay) 2(yk+ yy'ar(a; aa'yi )]a; aayi) ∑[2a4(1+2ay)-2ya1(1+2ay ad yy)-2n(ou 2y:ar(a;+ aayi) 11