Ta có:
\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(a^2b+ab^2=ab\left(a+b\right)\)
=> Ta cần chứng minh: \(a^2-ab+b^2\ge ab\) hay \(a^2+b^2\ge2ab\)
Ta có: \(a^2+b^2-2ab=a^2-2ab+b^2=\left(a-b\right)^2\ge0\)
\(\Rightarrow a^2+b^2-2ab\ge0\Rightarrow a^2+b^2\ge2ab\)
\(\Rightarrow a^2-ab+b^2\ge ab\)
\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3\ge a^2b+ab^2\)
Vậy \(\Rightarrow a^3+b^3\ge a^2b+ab^2\)