(a^4+b^4)/ab(a^3+b^3)
=(a^4+b^4)/(a^4b+ab^4)
因为:
(a^4+b^4)(a+b)=(a^5+b^5)+(a^4b+ab^4)
又(a^5+b^5)-(a^4b+ab^4)
=(a^4-b^4)(a-b)=(a-b)^2*(a^2+b^2)(a+b)>=0
所以(a^5+b^5)>=(a^4b+ab^4)
(a^4+b^4)(a+b)>=2(a^4b+ab^4)=2ab(a^3+b^3)
(a^4+b^4)/ab(a^3+b^3)>=2/(a+b)
同样,有:
(b^4+c^4)/bc(b^3+c^3)>=2/(b+c)
(c^4+a^4)/ca(c^3+a^3)>=2/(c+a)
(a^4+b^4)/ab(a^3+b^3)+(b^4+c^4)/bc(b^3+c^3)+(c^4+a^4)/ca(c^3+a^3)
=2[1/(a+b)+1/(b+c)+1/(c+a)]
设a>=b>=c
则2[1/(a+b)+1/(b+c)+1/(c+a)]
>=2[1/(2b)+1/(2c)+1/(2c)]=(1/b+2/c)
因为ab+bc+ac=abc
1/c+1/b+1/a=1
1/b+2/c=1/b+1/c+1/c=1-1/a+1/c>=1
所以式子>=1