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.

Actually both inequalities follow directly form Holder's Inequality. Obviously $q=\frac 32$ and $p = 3$ satisfy the condition $\frac 1p + \frac 1q = 1$. Then we have:
$$(1^q + 1^q)^{\frac 1q}(a^p + b^p)^{\frac 1p} \ge (a+b)$$
Substitute and we have:
$$(1^{\frac 32} + 1^{\frac 32})^{\frac 23}(a^3 + b^3)^{\frac 13} \ge (a+b)$$
Cube both sides and we have:
$$(1 + 1)^2(a^3 + b^3) \ge (a+b)^3$$
$$4(a^3 + b^3) \ge (a+b)^3$$
I'll leave the second inequality to you, but it's as easy as the first one.
