Power-mean Inequality: Prove that for positive reals $a$, $b$, $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$. 
Taiwan Quiz 2014:
Prove that for positive reals $a$, $b$, $c$ we have
$$3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$

Applying Power-mean inequality (weighted) in $$a_1={abc} , a_2={\frac{a^3+b^3+c^3}{3}}$$ with $r=1$, $s=\dfrac{1}{3}$ and weights $\dfrac{8}{9}, \dfrac{1}{9}$ respectively, we get
\begin{align}
  P(1) &\ge P\left(\frac{1}{3} \right) \\
 \frac{8}{9}\cdot {abc} +\frac{1}{9}\cdot {\frac{a^3+b^3+c^3}{3}} & \ge
 \left(
   \frac{8}{9}\sqrt[3]{abc}+
   \frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}
  \right)^3 \tag{1}
\end{align}
In the book it says this simplifies into
$$a^3 +b^3 +c^3 +24abc \le (a+b+c)^3$$
which is true by $AM \ge GM$, we get on $2$ terms
$$8abc\le(a+b)(b+c)(c+a) \tag{2}$$
Can someone explain me how the simplification happened from $(1)$ to $(2)$?

From (1), we have
$$27(24abc + a^3 + b^3 + c^3) \ge \left(8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right)^3.$$
Thus, it suffices to prove that
$$[3(a+b+c)]^3 \ge 27(24abc + a^3 + b^3 + c^3)$$
or
$$(a+b+c)^3 \ge 24abc + a^3 + b^3 + c^3.$$
