Question

a b∈r＋且a≠b 求证a^3+b^3>a^2b+ab^2

Answer 1

a^3+b^3-a^2b-ab^2
=a^2(a-b)+b^2(b-a)
=a^2(a-b)-b^2(a-b)
=(a-b)(a^2-b^2)
=(a-b)^2(a+b)
因为a b∈r＋且a≠b
所以a+b>0
(a-b)^2>0
所以
a^3+b^3-a^2b-ab^2>0
a^3+b^3>a^2b+ab^2

Answer 2

证明：欲证a^2/3+b^2/3>c^2/3
即证a^2+3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)+b^2＞c^2
只需证（a+b)^2-2ab+3a^(4/3)b^(2/3)
+3a^(2/3)b^(4/3)+b^2＞c^2
∵a+b=c,∴(a+b)^2=c^2
只需证
3a^(2/3)b^(2/3)[a^(2/3)+b^(2/3)]＞2ab
只需证a^(2/3)+b^(2/3)＞2/3*a^(1/3)b^(1/3)
∵a^(2/3)+b^(2/3)≥2a^(1/3)b^(1/3),
∴a^(2/3)+b^(2/3)＞2/3*a^(1/3)b^(1/3)成立,原不等式得证.

追问

求证a^3+b^3>a^2b+ab^2 看清题 不是2/3