Dimostrare $a^3+b^3\ge a^2b+ab^2$ per $a+b\ge 0$

Text

Prove that, for every pair of real numbers with $a+b\ge 0$, $a^3+b^3\ge a^2b+ab^2.$

Solution

Move everything to the left and factor: $a^3+b^3-a^2b-ab^2=a^2(a-b)-b^2(a-b)=(a-b)(a^2-b^2)=(a-b)^2(a+b).$ The factor $(a-b)^2\ge 0$ and, by hypothesis, $(a+b)\ge 0$; hence the product is $\ge 0$. Equality holds if $a=b$ or $a+b=0$. $\boxed{a^3+b^3\ge a^2b+ab^2\quad(a+b\ge 0).}$