66th Belarusian Mathematical Olympiad

Let $a_0,b_0,c_0$ be the initial numbers on the blackboard, and $a_i,b_i,c_i$ be the numbers written on the blackboard after the $i$-th move; let $S(a_0,b_0,c_0)$ and $S(a_i,b_i,c_i)$ be the sums of these numbers, respectively. Then after the $i+1$-th move we obtain the following numbers $$a_{i+1}=\frac{b_i^2+c_i^2}{a_i},b_{i+1}=\frac{c_i^2+a_i^2}{b_i},c_{i+1}=\frac{a_i^2+b_i^2}{c_i}.$$ We have the lower estimate for $S(a_{i+1},b_{i+1},c_{i+1})$: $$\begin{array}{rl}S(a_{i+1},b_{i+1},c_{i+1})& =\frac{b_i^2+c_i^2}{a_i}+\frac{c_i^2+a_i^2}{b_i}+\frac{a_i^2+b_i^2}{c_i}\ge \frac{2b_i{c}_i}{a_i}+\frac{2c_i{a}_i}{b_i}+\frac{2a_i{b}_i}{c_i}=\\ & =\frac{b_i{c}_i}{a_i}+\frac{b_i{c}_i}{a_i}+\frac{c_i{a}_i}{b_i}+\frac{c_i{a}_i}{b_i}+\frac{a_i{b}_i}{c_i}+\frac{a_i{b}_i}{c_i}=\\ & =\left(\frac{b_i{c}_i}{a_i}+\frac{a_i{b}_i}{c_i}\right)+\left(\frac{b_i{c}_i}{a_i}+\frac{c_i{a}_i}{b_i}\right)+\left(\frac{a_i{b}_i}{c_i}+\frac{c_i{a}_i}{b_i}\right)=\\ & =b_i\left(\frac{c_i}{a_i}+\frac{a_i}{c_i}\right)+c_i\left(\frac{b_i}{a_i}+\frac{a_i}{b_i}\right)+a_i\left(\frac{b_i}{c_i}+\frac{c_i}{b_i}\right)\ge \left[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\ge 2\right]\ge \\ & \ge 2(a_i+b_i+c_i)=2S(a_i,b_i,c_i).\end{array}$$ Therefore, $S(a_i,b_i,c_i)\le \frac{1}{2}S(a_{i+1},b_{i+1},c_{i+1})$ for all $i=0,1,2,3,4$. Hence, $$\begin{array}{rl}S(a_0,b_0,c_0)& \le \frac{1}{2}S(a_1,b_1,c_1)\le \frac{1}{2^2}S(a_2,b_2,c_2)\le \cdots \le \frac{1}{2^5}S(a_5,b_5,c_5)=\\ & =\frac{2016}{32}=63.\end{array}$$ Thus, the sum of the initial numbers on the blackboard is less than or equal to 63. The following example shows that the sum of the initial numbers on the blackboard can be equal to 63. Let $a_0=b_0=c_0=21$, then $$\begin{array}{rlrl}(a_0,b_0,c_0)=(21,21,21)& ⟶(42,42,42)=(2\cdot 21,2\cdot 21,2\cdot 21)& ⟶& \\ & ⟶(2^2\cdot 21,2^2\cdot 21,2^2\cdot 21)& ⟶\dots & ⟶(2^5\cdot 21,2^5\cdot 21,2^5\cdot 21)=(a_5,b_5,c_5),\end{array}$$ and the sum $$S(a_5,b_5,c_5)=S(2^5\cdot 21,2^5\cdot 21,2^5\cdot 21)=2^{5}S(21,21,21)=32\cdot 63=2016.$$