If we have $ r$ and write a pair of positive real numbers $ a$ and $ b$ satisfying the condition on the board, we prove that $ \frac{1}{a^2}+\frac{1}{b^2} \ge \frac{1}{r^2}$ $ \frac{1}{a^2}+\frac{1}{b^2} = \frac{a^2+b^2}{a^2b^2} \ge \frac{2ab}{a^2b^2}= \frac{2}{ab} = \frac {1}{r^2}$ qed If $ a_i = min a_k$ for all $ k$ then $ n\frac{1}{a_i^2}\ge \frac{1}{a_1^2}+...+\frac{1}{a_n^2}\ge \frac {1}{r^2}$ $ n\frac{1}{a_i^2}\ge \frac {1}{r^2}$ $ nr^2 \ge a_i^2$ $ \sqrt{n} r\ge a_i$ We have $ n=k^2$ $ rk\ge a_i$ qed

Surprisingly, induction works. Furthermore, we can strengthen it to if you apply the operation $ n$ times, there exists a number that does not exceed $ \sqrt{n+1}r$. To do this, we apply induction to the above statement. For $ n=0$ it is trivial, $ n=1$ follows from the fact that the geometric mean of $ a,b$, which is $ \sqrt{ab}$ is larger than the smaller of the two, so the smaller of the two cannot exceed $ \sqrt{ab}=\sqrt{2}r$. Now, assume the result holds true for $ n=k$. We will show the result to be true for $ n=k+1$. If we split the original number on the board $ r$ into $ a$ and $ b$, then we consider each of the numbers formed from $ a$, and each of the numbers formed from $ b$. If $ p$ of the $ k$ remaining operations were done to $ a$, and the remaining $ q$ operations were done to $ b$, then the induction hypothesis guarantees a number that does not exceed $ min(a\sqrt{p+1},b\sqrt{q+1})$ with $ p+q=k$, and $ ab=2r^2$. Assume the contrary, so that $ a\sqrt{p+1} > (\sqrt{k+2})r$ and $ b\sqrt{q+1} > (\sqrt{k+2})r$ $ p+1 > (k+2)\frac{r^2}{a^2}$, $ q+1>(k+2)\frac{r^2}{b^2}$ Summing, we get $ 1> r^2(\frac{1}{a^2}+\frac{1}{b^2})$ $ 2> ab(\frac{1}{a^2}+\frac{1}{b^2})$ $ 2>\frac{a}{b}+\frac{b}{a}$, contradicting AM-GM on the L.H.S. which gives $ \frac{a}{b}+\frac{b}{a} \ge 2$. Thus, we must have had $ min(a\sqrt{p+1},b\sqrt{q+1}) \le \(sqrt{k+2})r$, and the induction is complete. Of course, this may just look like the way of Russian_Man, but if you were not aware of that invariant, then this method would be the next most logical. Cheers, Rofler

Monovariant on $\sum_{\text{current numbers on the board}} \frac{1}{k^2}$. It is easy to see that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{a^2} + \frac{a^2}{4r^4} \ge \frac{1}{r^2}$ If all numbers are larger than $kr$, we would have the sum of $\frac{1}{i^2}$ less than $\frac{1}{r^2}$, which is a contradiction.

It is clear that at each step we should increase the minimum : When applying the process we have : $\min(a,b)\le \sqrt{2}r$ note that $\min(i+1)=\min(i)$ or $\min(i+1)\le \sqrt{2}\min(i)$ . Now analyzing the pattern we see that the best algorithm to increase the minimum is : $(r)\rightarrow (\sqrt{2}r,\sqrt{2}r)\rightarrow (2r,2r,\sqrt{2}r)\rightarrow (2r,2r,2r,2r).....$ which means we have : $\min(2^t-1)\le \sqrt{2}^t$ and $\min(i)\le \sqrt{2}^{t-1}$ where $i<2^t-1$ So : let $t\in\mathbb{Z}$ such that : $2^t-1\le k^2-1<2^{t+1}-1$ $\implies \sqrt{2}^t\le k<\sqrt{2}^{t+1}\\ \implies \min(k^2-1)\le \sqrt{2}^t \le k$

We define, for any positive real $x$ written on the board, a "level" $l$ defined as follows. The starting number $r$ has level $0$. Also, if a number $x$ with level $l_x$ is split into $a$ and $b$, then $a$ and $b$ both have level $l_a=l_b=l_x+1$. It's easy to see that $$\sum_{x \text{ on blackboard}} \frac{1}{2^{l_x}} = 1$$ For a number $x$ with level $l$, define $$f(x,l)= \frac{\log_2 x}{2^l} - \frac{l}{2^{l+1}}$$ A simple computation yields $f(x,l_x)=f(a,l_a)+(b,l_b)$, where $x$ with level $l_x$ is split into $a$ and $b$ with level $l_a=l_b=l_x+1$. Therefore, if the $n$ numbers in the end are $a_1,a_2,\cdots a_{n}$ with levels $l_1, l_2, \cdots l_n$, then $\sum_{i=1}^n \frac{1}{2^{l_i}}=1$, and $$\sum_{i=1}^n f(a_i,l_i) = f(r,0)=\log_2 r$$ Suppose $a_i > \sqrt{n}r$ for all $i$. Then $$\log_2 r > \sum_{i=1}^n \frac{\log_2 r\sqrt{n}}{2^{l_i}} - \sum_{i=1}^n \frac{l_i}{2^{l_i+1}}$$which simplifies to $$\sum_{i=1}^n \frac{l_i}{2^{l_i}} > \log_2 n$$ However, the function $g(x)=-x\ln x$ is concave on positive reals, so $$\sum_{i=1}^n g\left ( \frac{1}{2^{l_i}} \right ) \le n \cdot g\left ( \frac{1}{n} \cdot \sum_{i=1}^n \frac{1}{2^{l_i}} \right )$$which gives $$\sum_{i=1}^n \frac{l_i}{2^{l_i}} \le \log_2 n,$$a contradiction.

Solution from Twitch Solves ISL: The problem follows from one observation: Claim: [Main claim] The sum of the squares of reciprocals does not change under this operation. Proof. Clear. $\blacksquare$ Hence, if $x_1$, \dots, $x_n$ are the numbers on the board, when $n = k^2$, they satisfy \[ \frac{1}{r^2} = \sum \frac{1}{x_n^2} \]which implies $\min(x_1, \dots, x_n) \le {\sqrt n}r = kr$.

I proved that the sum of the reciprocals in the $i-$th operation is always stricly bigger than $\sqrt(i)/r$. This clearly implies the result, as if all the numbers were bigger than $kr$, then the sum of the reciprocals would be less than $k^2/rk=k/r$. We prove this by induction on the operation we're at: In the first operation this is true as $1/r \ge 1/r$. Now, if $S$ is this sum in the $i-$th operation, then $S \ge \sqrt(i)/r$ and we want to prove that $S'$, which is the sum in the next operation, is bigger than $\sqrt(i+1)/r$. Notice that $S'=S-1/t+1/a+1/b$, where $t$ is the erased number. But notice: $$ 1/a+1/b=a+b/ab=a+b/2t^2\ge \sqrt(ab)/t^2=t.\sqrt(2)/t^2=\sqrt(2)/t$$. Thus $S' \ge S-1/t+\sqrt(2)/t \ge \frac{\sqrt(i)+\sqrt(2)-1}{t}$. Now, we prove the LHS is bigger than $\sqrt(i+1)/t$. This occurs iff $\sqrt(i)+\sqrt(2)-1 > \sqrt(i+1)$, which occurs iff (by manipulating) $\sqrt(i^2+i) > \sqrt(2)-1$, which is true since $\sqrt(i^2+i) \ge \sqrt(2)$.

WLOG $r=1$ since we can just scale. Assign each number on the blackboard a weight such that the initial 1 starts with weight 1, and when a number with weight $w$ gets split into two numbers, each of the new numbers has weight $w/2$. Define the value of the configuration as the product of $$(number)^{weight}$$over all numbers on the board. Note that the sum of weights is always 1. It suffices to show that after $k^2-1$ moves, the value of the configuration is at most $k$. When a number $m$ with weight $w$ is split into $a$ and $b$, the value of the configuration is multiplied by $$\frac{(ab)^{w/2}}{m^w}=\frac{(2m^2)^{w/2}}{m^w}=2^{w/2}.$$In particular, this does not depend on $m$, $a$, or $b.$ Define $$a_2=1/2,a_3=a_4=1/4,a_5=a_6=a_7=a_8=1/8\cdots$$Now, define $$S_k=\sum_{i=2}^k a_k.$$ Claim 1: $$S_{k^2}\leq \log_2(k).$$This is equivalent to $$S_{k}\leq \frac{1}{2}\log_2(k).$$Note that when $k$ is a power of 2, equality holds. However, consider $k$ going up from one power of 2 to the next. The LHS will increase linearly by the same amount each time during this interval, while the RHS will increase concavely, and at the end of the interval, they become equal again. Therefore, the left side is at most the right side. Since splitting a number with weight $w$ multiplies the value of the configuration by $2^{w/2}$, after $m$ moves, the total value is at most $$2^{S_{m+1}}.$$Therefore, it suffices to show that $$2^{S_{k^2}}\leq k,$$which is equivalent to Claim 1, so we are done.

Claim : $\sum {\frac{1}{a_i^{2}}}$ is monovariant $\textbf{proof:}$ $$\frac{1}{a^2}+\frac{1}{b^2} = \frac{a^2+b^2}{a^2b^2} \ge \frac{2ab}{a^2b^2}= \frac{2}{ab} = \frac {1}{r^2}$$ Hence $$k^2\frac{1}{(min a_i)^2}\ge \sum_{i=1}^{k^2} \frac{1}{a_i^2} \ge \frac {1}{r^2} \implies kr \ge min a_i \blacksquare$$