Find the maximum of $\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$ for nonnegative $a$, $b$, $c$. 
Show that the maximum of
$$\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$$
is $4$ for nonnegative $a$, $b$, $c$.

An elegant elementary solution is preferred.
Generally, is there an easy way to show that
$$
\frac{
  \left( a_1 + \dots + a_n \right)^n - n^n \, a_1 \dots a_n
}
{
  a_1^n + \dots + a_n^n - n \, a_1 \dots a_n
}
\le
\left( n-1 \right)^{n-1},
$$
for nonnegative $a_1, \dots, a_n$? In other words,
$$
r_n(a_1, \dots, a_n)
\equiv
\frac{
  \dfrac{a_1^n+\dots+a_n^n}{n} - \,a_1 \dots a_n
}
{
  \left(\dfrac{ a_1+\dots+a_n }{n} \right)^n - a_1 \dots a_n
}
\ge
\left( \frac{n}{n-1} \right)^{n-1}.
$$
This would imply that
$$
\lim_{n \rightarrow \infty} r_n(a_1, \dots, a_n)
\ge e.
$$

Not a very elegant solution, I will admit.
We want to show that
$$ \frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}\leq 4$$
which is equivalent to (since the denominator is non-negative by AM-GM)
$$(a+b+c)^3-27abc \leq 4(a^3+b^3+c^3) - 12abc$$
or
$$a^3 + b^3 + c^3 + 3a^2(b+c) + 3b^2(c+a)+3c^2(a+b) + 6abc - 27abc \leq 4(a^3+b^3+c^3)-12abc$$
which further simplifies to
$$ a^2(b+c) + b^2(c+a) + c^2(a+b) \leq a^3 + b^3 + c^3 + 3abc$$
which in turn is equivalent to
$$ a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \geq 0$$
which is just Schur's inequality.
