, 2011

If $x$ and $z$ have opposite signs, or if one of them is $0$, then the inequality is obviously true, so $x$, $y$ and $z$ are either all positive or all negative. If they are all negative, we may replace $x$, $y$ and $z$ with $-x$, $-y$ and $-z$ without changing the conditions of the problem, so we can assume that $x$, $y$ and $z$ are all positive.

In that case, $0 < xy \le xz \le yz$, and so $1 = xy + yz + xz \ge xy + 2xz > 2xz$ which yields $xz < \frac{1}{2}$.

Setting $x = y = \frac{1}{n}$ for some natural number $n > 2$, the equation $xy + yz + xz = 1$ implies that $z = \frac{n^2-1}{2n} > \frac{1}{n}$ if $n > 2$. Hence $xz = \frac{n^2-1}{2n^2} = \frac{1}{2} - \frac{1}{2n^2}$, which can be made arbitrarily close to $\frac{1}{2}$. This shows that $\frac{1}{2}$ cannot be replaced by any smaller number.