Question

设a不等于b,比较a的平方（a+1)+b的平方（b+1）与a(a的平方+b）+b(b的平方+a)的大小

Answer 1

a2(a+1)+b2(b+1)-[a(a2+b)+b(b2+a)]=(a-b)2>0

so,a2(a+1)+b2(b+1)>a(a2+b)+b(b2+a)

number"2" represents for square.

PS:Right now I'm in the USA and could not type Chinese.

Hope you can understand.