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.

Solution 1:

(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 $9(a^3+b^3+c^3)\ge(a+b+c)^3$.

Solution 2:

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.