Initially there are $n≥3$ numbers $1$ on the board. At each turn, someone selects any $3$ numbers $x, y, z$ on the board, erases the three numbers, then writes the number $\frac {x+2y}{4}$ and $\frac {2y+z}{4}$ on board. Someone does $(n − 2)$ turns so that there are only $2$ numbers left on the board, namely $a$ and $b$. Prove that $ab> \frac {a+b}{n}$.
Can someone prove this? how to do this problem? Thank you
We show that the sum of reciprocals of the numbers on the board is strictly non-increasing. This follows by $$\frac{4}{x + 2y} \leq \frac{1}{x} + \frac{1}{2y},$$ $$\frac{4}{z + 2y} \leq \frac{1}{z} + \frac{1}{2y}$$ both of which are corollaries of Cauchy-Schwarz.
Thus we have $$\frac{1}{a} + \frac{1}{b} \leq n \cdot \frac{1}{1} = n$$ as desired.