Hello ~ ~ This question is actually not difficult to think about, but the calculation is more complicated.

The comparative questions are generally poor or business, obviously this problem is poor.

a^2b+b^2c+c^2a-ab^2-bc^2-ca^2

=a^2(b-c)+a(c^2-b^2)+bc(b-c)

=a^2(b-c)-(ab+ac)(b-c)+bc(b-c)

=(b-c)(a^2-ac-ab+bc)

=(b-c)[a(a-c)-b(a-c)]

=(b-c)(a-b)(a-c)

Because of a>b>c, B-C > 0,

a-b & gt； 0,

a-c & gt； 0,

So (b-c)(a-b)(a-c)>0,

That is, A 2B+B 2C+C 2A-AB 2-BC 2-CA 2 > 0,

a^2b+b^2c+c^2a>； ab^2+bc^2+ca^2