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 $B\ne \varnothing$. 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 $$a_i^{\prime }=\{\begin{array}{ll}0,& \text{if }{a}_i=0\text{ or }i=k\\ a_i+\frac{a_k}{|A|-1},& \text{if }{a}_i>0\text{ and }i\ne k\end{array}$$ 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 $(a_{k-1}^{\prime },a_k^{\prime })=(a_{k-1}+a_k,0)$. 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 $|a_{k-2}-a_{k-1}^{\prime }|=a_{k-1}+a_k$; for $i=k-1$ increases from $|a_{k-1}-a_k|=a_k-a_{k-1}$ to $|a_{k-1}^{\prime }-a_k^{\prime }|=a_{k-1}+a_k$; for $i=k$ increases from $|a_k-a_{k+1}|=a_{k+1}-a_k$ to $|a_k^{\prime }-a_{k+1}|=a_{k+1}$.

$a_{k-1}>a_k>a_{k+1}$. 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 $(a_{k-1}^{\prime },a_k^{\prime },a_{k+1}^{\prime })=(a_{k-1}+a_k/2,0,a_{k+1}+a_k/2)$. 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 $|(a_{k-2}-a_{k-1}^{\prime })(a_{k-1}^{\prime }-a_k^{\prime })(a_k^{\prime }-a_{k+1}^{\prime })(a_{k+1}^{\prime }-a_{k+2})|=(a_{k-1}+a_k/2{)}^2(a_{k+1}+a_k/2{)}^2$. But $$\begin{array}{rl}& (a_{k-1}+a_k/2{)}^2(a_{k+1}+a_k/2{)}^2\\ & =(a_{k-1}(a_{k-1}+a_k)+a_k^2/4)(a_{k+1}(a_{k+1}+a_k)+a_k^2/4)\\ & >a_{k-1}(a_k+a_{k-1})(a_k+a_{k+1})a_{k+1}\\ & >a_{k-1}(a_k-a_{k-1})(a_k-a_{k+1})a_{k+1}\end{array}$$

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, $$\begin{array}{rl}{a}_1{a}_{2011}(a_1-a_{2011})& =\frac{1}{\alpha \beta }(\alpha a_1)(\beta a_{2011})(a_1-a_{2011})\\ & \le \frac{1}{\alpha \beta }{\left(\frac{\alpha a_1+\beta a_{2011}+(a_1-a_{2011})}{3}\right)}^3\\ & =\frac{1}{\alpha \beta }{\left(\frac{(\alpha +1)a_1+(\beta -1)a_{2011}}{3}\right)}^3\end{array}$$ So we choose $\alpha$ and $\beta$ such that we obtain $s$ in the end, that is, $\alpha +1=\beta -1\iff \beta -\alpha =2$; the equality can occur, that is, $\alpha a_1=\beta a_{2011}=a_1-a_{2011}\iff a_{2011}=(1-\alpha )a_1$ and $a_1=(\beta +1)a_{2011}$, that is, $1=(1-\alpha )(\beta +1)\iff -\alpha \beta =\alpha -\beta =-2$. 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 $$\begin{array}{rl}{a}_1{a}_{2011}(a_1-a_{2011})& \le \frac{1}{\alpha \beta }{\left(\frac{(\alpha +1)a_1+(\beta -1)a_{2011}}{3}\right)}^3\\ & =\frac{1}{2}{\left(\frac{\sqrt{3}(a_1+a_{2011})}{3}\right)}^3=\frac{\sqrt{3}}{18}{s}^3\end{array}$$

Now we optimize the rest. If $a_3+a_5+\cdots +a_{2009}=\frac{2011}{2}-s$, $$a_3^2{a}_5^2\dots a_{2009}^2\le {\left(\frac{a_3+a_5+\cdots +a_{2009}}{1004}\right)}^{2008}={\left(\frac{\frac{2011}{2}-s}{1004}\right)}^{2008}$$

$$\begin{array}{rl}E& =a_1{a}_{2011}|a_1-a_{2011}|a_3^2{a}_5^2\dots a_{2009}^2\\ & \le \frac{\sqrt{3}}{18}{s}^3\cdot {\left(\frac{\frac{2011}{2}-s}{1004}\right)}^{2008}=\frac{\sqrt{3}}{18{\gamma }^3}(\gamma s{)}^3{\left(\frac{\frac{2011}{2}-s}{1004}\right)}^{2008}\\ & \le \frac{\sqrt{3}}{18{\gamma }^3}{\left(\frac{3\gamma s+2008\cdot \frac{\frac{2011}{2}-s}{1004}}{2011}\right)}^{2011}\\ & =\frac{\sqrt{3}}{18{\gamma }^3}{\left(\frac{2011+(3\gamma -2)s}{2011}\right)}^{2011}\end{array}$$