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

Question

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 $$ \lim_{n \rightarrow \infty} r_n(a_1, \dots, a_n) \ge e. $$

Answer 1

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.

Answer 2

Here is an analytical proof for the general case, although I suspect better solutions exist. The $n = 2$ case is obvious, and we shall assume $n \ge 3$.

Without loss of generality, we can assume $a_1 \le a_2 \le \dots \le a_n$. First, since the function $f(x) = x^n$ is convex, we have $$ a_2^n + \dots + a_n^n \ge (n - 1) \left( \frac{ a_2 + \dots + a_n } { n - 1 } \right)^n, $$ with the equality achieved at $a_2 = \dots = a_n$.

1) If $a_1 = 0$, then $a_1 \cdots a_n = 0$, we have the \begin{align} R &\equiv \frac{ \left( a_1 + a_2 + \dots + a_n \right)^n - n^n \, a_1 \cdots a_n }{ a_1^n + a_2^n + \dots + a_n^n - n \, a_1 \cdots a_n } \\ &= \frac{ \left( a_2 + \dots + a_n \right)^n }{ a_2^n + \dots + a_n^n } \le (n - 1)^{n - 1}. \end{align}

2) If $a_1 > 0$, all other $a_k > 0$. The inequality makes sense only if at least two $a_k$ are different. Thus, we can assume $a_n > a_1$. Now suppose for some $m$ ($1 < m < n$), we have $a_1 \le a_m < a_n$. Consider the following infinitesimal adjustment \begin{align} a_1 &\rightarrow a_1 - (a_n - a_m) \, a_1 \, dt \\ a_m &\rightarrow a_m + (a_n - a_1) \, a_m \, dt \\ a_n &\rightarrow a_n - (a_m - a_1) \, a_n \, dt. \end{align} It leaves $a_1 + \dots + a_n$ and $a_1 \cdots a_n$ invariant, and changes $a_1^n + \dots + a_n^n$ by \begin{align} d (a_1^n + \dots + a_n^n) &= d a_1^n + d a_m^n + d a_n^n \\ &= n \, \left[ -(a_n - a_m) \, a_1^n +(a_n - a_1) \, a_m^n -(a_m - a_1) \, a_n^n \right] \, dt \\ &= n \, \left[ (a_n - a_m) (a_m^n - a_1^n) -(a_m - a_1) (a_n^n - a_m^n) \right] \, dt \\ &= \big[ (a_m^{n-1} + a_m^{n-2} a_1 + \cdots + a_1^{n-1}) \\ &\; -(a_m^{n-1} + a_m^{n-2} a_n + \cdots + a_n^{n-1}) \big] \, n \, (a_n - a_m) (a_m - a_1)\, dt \\ &< 0. \end{align} The last expression is less than zero, because $a_1 < a_n$. This means that our adjustment always increases the target function $R$, and since it decreases $a_1$, we should set $a_1$ to the smallest possible value, which is $0$. This is the case 1).

3) Last, if no such $m$ exists, then for $m > 1$, $a_m = a_n$. Then, we can choose $m = 2$ and use the same transform, which will create some difference between $a_n$ and $a_m$, and return to case 2). This completes the proof.