lagrange恒等式证明.用和式的恒等变形.

Question

lagrange恒等式证明.
用和式的恒等变形.

Answer 1

一个推论,利用拉格朗日恒等式可以证明柯西不等式,好了,下面开始给你证明.‘
有一个适合中学生的拉格朗日恒等式:
[(a1)^2+(a2)^2][(b1)^2+(b2)^2]=
[(a1)(b1)+(a2)(b2)]^2+[(a2)(b1)-(a1)(b2)]^2
[(a1)^2+(a2)^2+(a3)^2][(b1)^2+(b2)^2+(b3)^2]=
=[(a1)(b1)+(a2)(b2))+(a3)(b3)]^2+[(a2)(b1)-(a1)(b2)]^2+
+[(a3)(b1)-(a1)(b3)]^2+[(a2)(b3)-(a3)(b2)]^2
[(a1)^2+...+(an)^2][(b1)^2+...+(bn)^2]=
=[(a1)(b1)+...+(an)(bn)]^2+[(a2)(b1)-(a1)(b2)]^2+
+[(a3)(b1)-(a1)(b3)]^2+..+[(a(n-1))(bn)-(an)(b(n-1))]^2
用数学归纳法证明.
1.
显然n=1时,[(a1)^2][(b1)^2]=[(a1)(b1)]^2.
拉格朗日恒等式成立.
2.
设n=k时,拉格朗日恒等式成立.
当n=k+1时,
[(a1)^2+...+(a(n+1))^2][(b1)^2+...+(b(n+1))^2]-
-[(a1)(b1)+...+(a(n+1))(b(n+1))]^2=
={[(a1)^2+...+(an)^2][(b1)^2+...+(bn)^2]-
-[(a1)(b1)+...+(an)(bn)]^2}+
+{[(a(n+1))^2(b1)^2+(b(n+1))^2(a1)^2]+..+
+[(a(n+1))^2(bn)^2+(b(n+1))^2(an)^2]-
-2a(n+1)b(n+1)[(a1)(b1)+...+(an)(bn)]}=
={[(a2)(b1)-(a1)(b2)]^2+[(a3)(b1)-(a1)(b3)]^2+..+
+[(a(n-1))(bn)-(an)(b(n-1))]^2}+
+{[(a(n+1))^2(b1)^2-2a(n+1)b(n+1)(a1)(b1)+
+(b(n+1))^2(a1)^2]+..+[(a(n+1))^2(bn)^2-
-2a(n+1)b(n+1)(an)(bn)+(b(n+1))^2(an)^2]}=
={[(a2)(b1)-(a1)(b2)]^2+[(a3)(b1)-(a1)(b3)]^2+..+
+[(a(n-1))(bn)-(an)(b(n-1))]^2}+
+{[(a(n+1))(b1)-b(n+1)(a1)]^2+
+..+[(a(n+1))(bn)-b(n+1)(bn)]^2}
所以n=k+1时,拉格朗日恒等式成立.
这样数学归纳法证明了拉格朗日恒等式.