Brazilian Math Olympiad

In what follows, indices are taken modulo $2011$ and $E = \left| \prod_{cyc} (a_n - a_{n+1}) \right|$.

Lemma. If $E$ is maximum, for every $i \in \{1, 2, \dots, 2011\}$, one of the numbers $a_{i-1}, a_i, a_{i+1}$ is zero.

Proof. Suppose, by means of contradiction, that $E$ is maximum and there exists $a_i$ such that $a_{i-1}, a_i, a_{i+1}$ are all nonzero (that is, $a_{i-1}a_i a_{i+1} > 0$). Define $A = \{a_i \mid a_i > 0\}$ and $B = \{a_i \mid a_{i-1}a_i a_{i+1} > 0\}$. Then $B \subset A$ and $B \neq \emptyset$. Let $a_k = \min B$ and consider $a_{k-1}$ and $a_{k+1}$. We have the following cases:

* $a_k < a_{k-1}$ and $a_k < a_{k+1}$. Let
$$a'_i = \begin{cases} 0, & \text{if } a_i = 0 \text{ or } i = k \\ a_i + \frac{a_k}{|A|-1}, & \text{if } a_i > 0 \text{ and } i \neq k \end{cases}$$
That is, we make $a_k$ be zero and distribute it among the remaining nonzero terms. So $|a_i - a_{i+1}|$ remains unchanged if $a_i, a_{i+1} \in A$ and $k \notin \{i, i+1\}$, or $a_i, a_{i+1} \notin A$; increases from $|a_i - a_{i+1}| = \max\{a_i, a_{i+1}\}$ to $\max\{a_i, a_{i+1}\} + \frac{a_k}{|A|-1}$ if $a_i \notin A$ or $a_{i+1} \notin A$, but not both; increases from $|a_{k \pm 1} - a_k| = a_{k \pm 1} - a_k$ to $a_{k \pm 1} + \frac{a_k}{|A|-1}$ if $k \in \{i, i+1\}$.

* $a_{k-1} < a_k < a_{k+1}$. This means that $a_{k-1} \notin B$, and $a_k \in B$, $a_{k-1} > 0$, that is, $a_{k-1} \in A \setminus B$, which means $a_{k-2} = 0$. In this case, we exchange $(a_{k-1}, a_k)$ for $(a'_{k-1}, a'_k) = (a_{k-1} + a_k, 0)$. Then $|a_i - a_{i+1}|$ remains unchanged for $i \notin \{k-2, k-1, k\}$; for $i = k-2$ increases from $|a_{k-2} - a_{k-1}| = a_{k-1}$ to $|a_{k-2} - a'_{k-1}| = a_{k-1} + a_k$; for $i = k-1$ increases from $|a_{k-1} - a_k| = a_k - a_{k-1}$ to $|a'_{k-1} - a'_k| = a_{k-1} + a_k$; for $i = k$ increases from $|a_k - a_{k+1}| = a_{k+1} - a_k$ to $|a'_{k} - a_{k+1}| = a_{k+1}$.

* $a_{k-1} > a_k > a_{k+1}$. Analogous to the previous case.

* $a_k > a_{k-1}$ and $a_k > a_{k+1}$. This means $a_{k-1}, a_{k+1} \in A \setminus B$, that is, $a_{k-2} = a_{k+2} = 0$. In this case, exchange $(a_{k-1}, a_k, a_{k+1})$ for $(a'_{k-1}, a'_k, a'_{k+1}) = (a_{k-1} + a_k/2, 0, a_{k+1} + a_k/2)$. All differences $|a_i - a_{i+1}|$ remain unchanged except if $i \in \{k-2, k-1, k, k+1\}$. The only change is $|(a_{k-2}-a_{k-1})(a_{k-1}-a_k)(a_k-a_{k+1})(a_{k+1}-a_{k+2})| = a_{k-1}(a_k-a_{k-1})(a_k-a_{k+1})a_{k+1}$ to $|(a_{k-2}-a'_{k-1})(a'_{k-1}-a'_k)(a'_k-a'_{k+1})(a'_{k+1}-a_{k+2})| = (a_{k-1}+a_k/2)^2(a_{k+1}+a_k/2)^2$. But
$$\begin{aligned} & (a_{k-1} + a_k/2)^2 (a_{k+1} + a_k/2)^2 \\ & = (a_{k-1}(a_{k-1} + a_k) + a_k^2/4)(a_{k+1}(a_{k+1} + a_k) + a_k^2/4) \\ & > a_{k-1}(a_k + a_{k-1})(a_k + a_{k+1})a_{k+1} \\ & > a_{k-1}(a_k - a_{k-1})(a_k - a_{k+1})a_{k+1} \end{aligned}$$

Now we only have groups with one or two consecutive nonzero variables. For a group $(0, a_k, 0)$, we obtain the product $|(a_{k-1} - a_k)(a_k - a_{k+1})| = a_k^2$; for a group $(0, a_k, a_{k+1}, 0)$, we obtain $|(a_{k-1} - a_k)(a_k - a_{k+1})(a_{k+1} - a_{k+2})| = a_k a_{k+1}|a_{k+1} - a_k|$. Notice that the groups can be interchanged, such that we can suppose wlog that all groups with two nonzero variables are contiguous.

Lemma. If $E$ is maximum then there is exactly one group with two nonzero variables.
Suppose, that there are at least two groups of nonzero variables $(0, a, b, 0)$ and $(0, c, d, 0)$. By the above remark, we can suppose wlog that the groups are consecutive, that is, it's $(0, a, b, 0, c, d, 0)$. Exchange these variables for $(0, a + b/2, 0, (b+c)/2, 0, d + c/2, 0)$. The product $abcd|(a-b)(c-d)|$ is exchanged for $(a+b/2)^2((b+c)/2)^2(d+c/2)^2$. But we already know that $(a+b/2)^2 > a|a-b|$, $(d+c/2)^2 > d|c-d|$ and, by AM-GM, $((b+c)/2)^2 \ge bc$. Multiplying everything yields the lemma.

Combining the two lemmas, we can suppose wlog that the nonzero variables are the ones with odd indices, that is, $a_1, a_3, \dots, a_{2011}$. In this case, we obtain the product $a_1 a_{2011} |a_1 - a_{2011}| a_3^2 a_5^2 \dots a_{2009}^2$, and we can optimize it locally.
Let $a_1 + a_{2011} = s$ and suppose wlog $a_1 > a_{2011}$. Let $\alpha, \beta$ be positive real numbers to be determined. By AM-GM,
$$\begin{aligned} a_1 a_{2011} (a_1 - a_{2011}) &= \frac{1}{\alpha\beta} (\alpha a_1) (\beta a_{2011}) (a_1 - a_{2011}) \\ &\le \frac{1}{\alpha\beta} \left( \frac{\alpha a_1 + \beta a_{2011} + (a_1 - a_{2011})}{3} \right)^3 \\ &= \frac{1}{\alpha\beta} \left( \frac{(\alpha + 1)a_1 + (\beta - 1)a_{2011}}{3} \right)^3 \end{aligned}$$
So we choose $\alpha$ and $\beta$ such that
* we obtain $s$ in the end, that is, $\alpha + 1 = \beta - 1 \Leftrightarrow \beta - \alpha = 2$;
* the equality can occur, that is, $\alpha a_1 = \beta a_{2011} = a_1 - a_{2011} \Leftrightarrow a_{2011} = (1-\alpha)a_1$ and $a_1 = (\beta+1)a_{2011}$, that is, $1 = (1-\alpha)(\beta+1) \Leftrightarrow -\alpha\beta = \alpha - \beta = -2$.
Thus $-\alpha$ and $\beta$ are the roots of the quadratic $t^2 - 2t - 2 = 0$. Hence $\alpha = \sqrt{3} - 1$ and $\beta = 1 + \sqrt{3}$, and
$$\begin{aligned} a_1 a_{2011} (a_1 - a_{2011}) &\le \frac{1}{\alpha\beta} \left( \frac{(\alpha + 1)a_1 + (\beta - 1)a_{2011}}{3} \right)^3 \\ &= \frac{1}{2} \left( \frac{\sqrt{3}(a_1 + a_{2011})}{3} \right)^3 = \frac{\sqrt{3}}{18} s^3 \end{aligned}$$

Now we optimize the rest. If $a_3 + a_5 + \dots + a_{2009} = \frac{2011}{2} - s$,
$$a_3^2 a_5^2 \dots a_{2009}^2 \le \left( \frac{a_3 + a_5 + \dots + a_{2009}}{1004} \right)^{2008} = \left( \frac{\frac{2011}{2} - s}{1004} \right)^{2008}$$

$$\begin{align*} E &= a_1 a_{2011} |a_1 - a_{2011}| a_3^2 a_5^2 \dots a_{2009}^2 \\ &\le \frac{\sqrt{3}}{18} s^3 \cdot \left(\frac{\frac{2011}{2}-s}{1004}\right)^{2008} = \frac{\sqrt{3}}{18\gamma^3} (\gamma s)^3 \left(\frac{\frac{2011}{2}-s}{1004}\right)^{2008} \\ &\le \frac{\sqrt{3}}{18\gamma^3} \left( \frac{3\gamma s + 2008 \cdot \frac{\frac{2011}{2}-s}{1004}}{2011} \right)^{2011} \\ &= \frac{\sqrt{3}}{18\gamma^3} \left( \frac{2011 + (3\gamma - 2)s}{2011} \right)^{2011} \end{align*}$$