Use Schur for $r=1$ to prove that product of all integers strictly decreases.

The above solution is too clever. Here is another solution

A very simple monovariant is the sum of the squares of all integers on the blackboard. 1. Since the sum $S$ of all integers on the blackboard never changes, the sum of all squares is at most $S^2$. 2. $(a + b - c)^2+ (b + c - a)^2+ (c + a - b)^2$ is at least $a^2+b^2+c^2$, with equality occurring only for $a=b=c$. The claimed inequality is equivalent to $(a-b)^2+(a-c)^2+(b-c)^2\ge0$.

One can either 1. Get an Upper Bound of the sum of squares and prove that it strictly increases every time. 2. Get an Lower Bound of the product and prove that it strictly decreases every time. 1 is easy inequality, 2 is Schur with $r=1$. Not too difficult.

We can just observe that we get a kind of majorized sequence of numbers in the end, id est, $a, b, c \cdots $ such that $c>a+b$ and so on.

What if the numbers are reals instead of integers?

I was stuck on this problem for a long time. Then, I gave up and wanted to check the solutions. Just at that moment, I realised the numbers on the board are positive integers and not real numbers and at that moment I understood why this problem was considered the easiest one of the MEMO... Solution 1 (Monovariant: Product of Numbers): Consider the product of all numbers on the board. Take any three numbers $a,b,c$ on the blackboard to make a move. We can write $a=x+y,b=y+z,c=z+x$ due to Ravi. Then \[ (a+b-c)(b+c-a)(c+a-b) = 8xyz \leq (x+y)(y+z)(z+x) = abc \]by AM-GM. Equality never arises, as then we'd have to have $x=y=z$ and thus $a=b=c$, which we cannot use. Therefore, the product is strictly decreasing and as all numbers on the blackboard are integers, the product also remains an integer. Thus, it decreases by at least $1$ each move and will therefore somewhen reach non-positivity, assuming there was an infinite sequence of moves. That is the desired contradiction. Solution 2: (Monovariant: Largest Number, Invariant: Sum of Numbers): We'll proceed by induction. Base case: For $n=3$: Take the numbers $a,b,c$. Let $a \geq b \geq c$. Then $a+b-c \geq a$. If $b \neq c$, then $a+b-c \geq a+1$. Else, we get the number $a,2b-a,a$. A move on that turn will give us $a+a-(2b-a)=3a-2b>a$, since we cannot have $a=b=c$. Thus the maximum is strictly increasing. But the sum of the numbers on the blackboard stay constant, say $S :=a+b+c$. Therefore, the maximum will somewhen reach $\tfrac{a+b+c}{2}$. But by the triangle inequality, the three numbers on the blackboard don't form a triangle anymore. Thus, we cannot have an infinite sequence of moves for $n=3$. Induction Hypothesis: There is no infinite sequence of moves for $n$ numbers on the blackboard. Induction Step: Assume otherwise. Let $a_1,a_2,\dots,a_{n+1}$ be the starting numbers. Then $S:=a_1+a_2+\dots+a_{n+1}$ is invariant. Now look at $a:=\max(a_1,a_2,\dots,a_{n+1})$. Either, somewhen, by doing moves, there will be a number larger than $a$, then the maximum increases. Or assume, $a$ would stay the largest. Somewhen, there has to be a move involving $a$, since there is no infinite sequence of moves with only the other $n$ numbers due to the Induction Hypothesis. Let $a \geq b \geq c$ be three numbers that represent the first move involving $a$. That gives us $a+b-c,b+c-a,c+a-b$. The only possibility that the maximum doesn't increase would be if $b=c$. That then produces two copies of $a$, though. By a similar argument, we will have a move involving $a$ somewhen again. If the maximum does not increase, we'd produce another copy of $a$. So the maximum has to increase some time, else we'd end up with $a_1=a_2=\dots=a_{n+1}=a$ with which the sequence of moves would terminate. Now as the maximum increases, that maximum somewhen reaches $\tfrac{S}{2}$. By the triangle inequality, the maximum then does not form a triangle with any two other numbers, thus the set of moves can then only involve the other $n$ numbers. But by the Induction Hypothesis, it'll terminate.

Note that since the sorted list of numbers increases in lexicographical order (post 4), considering this list as a single integer in base S (= sum of all numbers) also gives a monovariant.

It is a nice observation that if $a,b,c$ are positive real numbers, the statement of problem is still correct. For proof, just observe that sequence of moves cannot be infinite if $a,b,c$ are positive rationals (we just multiply all numbers with the lowest common denominator and work with integers) and then we just need to recall that there exist Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ so we can write each number as ordered $n$-tuple of rational numbers with standard coordinate addition. Now, because finite number of moves stabilizes each coordinate of each number and since number of coordinates is finite, we can deduce that each coordinate is stabilized after finite number of moves.

Very nice. Just let me note that the axiom of choice is overkill here, because the finitely many numbers generate a vector space of finite dimension. I have still not decided on the case of real numbers.

By AM-GM, the sum of squares strictly increasing(since equality achive in each step when all 3 numbers are equal which contradicts the condition of problem) and clearly, the sum of all numbers is invarant If this sequences of moves be infinite,then the sum of squares tends to infinity,while the sum stills be invarant ,which is impossible . I wonder if my solution is valid? Can anyone check please.

Yes, that is a correct solution.

the sum of the numbers on the board does not change with each move. If we look at the numbers a, b, c, at least one of them decreases with each move. So this process cannot continue indefinitely.