Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$ 
Let $a$, $b$ and $c$ be positive real numbers.
  $(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
  $(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$

For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab^2$ but I'm not sure how to prove it.

(i) By AM-GM, $a^3+a^3+b^3\ge3a^2b$ and $a^3+b^3+b^3\ge3ab^2$. 
Adding these inequalities gives $3a^3 + 3b^3 \ge 3a^2b + 3ab^2$, so $4(a^3+b^3)\ge(a+b)^3$.
(ii) Again by AM-GM, $a^3 + b^3 + c^3 \ge 3abc$. 
Add two times that to the six cyclic versions of $a^3 + a^3 + b^3 \ge 3a^2b$ to get $8(a^3 + b^3 + c^3)\ge 3a^2b + 3a^2c + 3b^2a+3b^2c+3c^2a+3c^2b+6abc$. 
So $9(a^3+b^3+c^3)\ge(a+b+c)^3$.
