Recently, I have found this problem:
Initially, only the integer $44$ is written on a board. An integer a on the board can be re- placed with four pairwise different integers $a_1, a_2, a_3, a_4$ such that the arithmetic mean $\frac{1}{4}(a_1+a_2+a_3+a_4)$ of the four new integers is equal to the number $a$. In a step we simultaneously replace all the integers on the board in the above way. After $30$ steps we end up with $n = 4^{30}$ integers $b_1, b_2, . . . , b_n$ on the board. Prove that: $$\frac{b_1^2+b_2^2+b_3^2+\cdots+b_n^2}{n}\geq2011$$
I've spent some days trying to solve this problem, but I've no idea of how to figure it out. Any idea?
We replace an integer $x$ by the $4$ distinct integers $x-a, x-b, x-c, x-d$ where $a+b+c+d = 0.$ Consider $$(x-a)^2+ (x-b)^2 +(x-c)^2+ (x-d)^2-4x^2=a^2+b^2+c^2+d^2.$$ This is minimized, with value $10$, when $$\{a,b,c,d\}=\{-1,-2,1,2\}$$
If you apply this observation to the problem, you will see that there is a minimum increase of $2.5$ at each step. After $30$ steps, the average of the squares must be at least $$44^2+30\cdot2.5=2011.$$