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 $9(a^3+b^3+c^3)\ge(a+b+c)^3$.
for (i); notice that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This is the standard factorization for the sum of cubes. Then $4(a^2-ab+b^2)\ge(a+b)^2$.
Thus, $$4a^2-4ab+4b^2\ge (a+b)^2$$ $$a^2+2ab+b^2+3a^2-6ab+3b^2\ge (a+b)^2$$ $$(a+b)^2+3(a-b)^2\ge(a+b)^2$$ Since $3(a-b)^2$ is always positive or $0$, the inequality is proven.
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.
