, 2008

Solution:
We'll prove it by induction.
For $n=1$ the desired result becomes $1 > x_{1}$ which is true.
Let the result be true for some natural number $n \geq 1$.
We'll prove it to be true for $n+1$ as well, and we'll be done.
So let $x_{1}, x_{2}, \ldots, x_{n}, x_{n+1}$ be $n+1$ given real numbers with $0 < x_{i} < 1$, for any $i$. We wish to show that
$$1 + \sum_{1 \leq i < j \leq n+1} x_{i} x_{j} > x_{1} + x_{2} + \ldots + x_{n} + x_{n+1}$$
If $1 \geq x_{1} + x_{2} + \ldots + x_{n}$ then we rewrite (5) equivalently as
$$1 - (x_{1} + x_{2} + \ldots + x_{n}) + x_{n+1}(x_{1} + x_{2} + \ldots + x_{n} - 1) + \sum_{1 \leq i < j \leq n} x_{i} x_{j} > 0$$
This is also written as
$$(1 - x_{n+1})\left[1 - (x_{1} + x_{2} + \ldots + x_{n})\right] + \sum_{1 \leq i < j \leq n} x_{i} x_{j} > 0$$
which is clearly true.
If instead $x_{1} + x_{2} + \ldots + x_{n} > 1$ then $x_{n+1}(x_{1} + x_{2} + \ldots + x_{n}) > x_{n+1}$, i.e.
$$x_{n+1} x_{1} + x_{n+1} x_{2} + \ldots + x_{n+1} x_{n} > x_{n+1}$$
By the induction hypothesis applied to the $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$ we also know that
$$1 + \sum_{1 \leq i < j \leq n} x_{i} x_{j} > \sum_{i=1}^{n} x_{i}$$
Adding (6) and (7) together we get the validity of (5) in this case too, and we are done.