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