Question

证明不等式(a^4+b^4)(a^2+b^2)>=(a^3+b^3)^2

Answer 1

(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2

=a^6+a^4b^2+a^2b^4+b^6-a^6-2a^3b^3-b^6

=a^4b^2+a^2b^4-2a^3b^3

=a^2b^2(a^2-2ab+b^2)

=a^2b^2(a-b)^2

所以若a=0或b=0或a=b

所以a^2b^2(a-b)^2=0

则(a^4+b^4)(a^2+b^2)=(a^3+b^3)^2

若a≠0且b≠0且a≠b

则a^2b^2(a-b)^2>0

则(a^4+b^4)(a^2+b^2)>(a^3+b^3)^2