Question

急!a1〈a2〈a3,b1〈b2〈b3,a1+a2+a3=b1+b2+b3,a1a2+a1a3+a2a3=b1b2+b1

Answer 1

证明：

a1^2+a2^2+a3^2=(a1+a2)^2-2a1a2+a3^3=(a1+a2+a3)^2-2(a1+a2)a3-2a1a2

=(a1+a2+a3)^-2（a1a3+a2a3+a1a2）

=(b1+b2+b3)^2-2(b1b2+b1b3+b2b3)

=(b1+b2)^2+2(b1+b2)b3+b3^2-2(b1b2+b1b3+b2b3)

=b1^2+2b1b2+b2^2+2b1b3+2b2b3+b3^2-2b1b2-2b1b3-2b2b3

=b1^2+b2^2+b3^2