<s>[INST] How to prove that $a^3 + b^3 \geq a^2b + ab^2$? Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--

For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$

I have no idea how to begin but this is what I did.
By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)
Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$
Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.
(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)
 [/INST] Hint:
$$a^3+b^3-(a^2b+ab^2)=(a+b)(a^2-ab+b^2)-ab(a+b)=(a+b)(a-b)^2\ge0$$ if $a+b\ge0$
</s>