(a^4 + b^4)-(ba^3 + ab^3)
=a^4-ba^3+b^4-ab^3
=a^3(a-b)+b^3(b-a)
=a^3(a-b)-b^3(a-b)
=(a-b)(a^3-b^3)
=(a-b)(a-b)(a^2+ab+b^2)
=(a-b)^2[(a+b/2)^2+3b^2/4]
因为a>0,b>0,
所以(a+b/2)^2>0,3b^2/4>0
所以
[(a+b/2)^2+3b^2/4]>0
又(a-b)^2≥0
所以(a-b)^2[(a+b/2)^2+3b^2/4]≥0
(a^4 + b^4)-(ba^3 + ab^3)≥0
所以
a^4 + b^4≥ba^3 + ab^3