Suppose that $ n \geq 2$ and $ x_1, x_2, \ldots, x_n$ are real numbers between 0 and 1 (inclusive). Prove that for some index $ i$ between $ 1$ and $ n - 1$ the
inequality
\[ x_i (1 - x_{i+1}) \geq \frac{1}{4} x_1 (1 - x_{n})\]
If
Suppose that $ x_{i}(1-x_{i+1})<\frac{1}{4}x_{1}(1-x_{n})\leq \frac{1}{4}$ is true for all $ i=1,n-1$
Then for $ i=1$ we get:
$ x_2>\frac{3}{4}>\frac{1}{2}$
Else, is obvious that if $ x_i>\frac{1}{2}$ then $ x_{i+1}>\frac{1}{2}$
because $ x_{i}(1-x_{i+1})<\frac{1}{4}$.
Then $ x_{n-1}>\frac{1}{2}$
(If $ x_n=1$ the problem is solved alone)
So $ x_{n-1}(1-x_n)<\frac{1}{4}x_{1}(1-x_{n})$
$ \implies x_{n-1}<\frac{1}{4}$ That is a contradiction.