Let $x,y,z$ be positive real numbers where $$ \frac{1}{3} \leq xy + yz + zx \leq 3. $$ Determine the range of values for $xyz$ and $x+y+z$.
I found this question on the British Mathematical Olympiad from 1993, but I can't seem to make any headway with it. Any ideas?
$3 \ge xy+yz+zx \ge 3\sqrt[3]{x^2y^2z^2} \implies xyz \le 1$. As $x=y=z=1$ gives equality, this is the maximum.
As $z \to 0$, we can still have $\frac13 \le xy+yz+zx \to xy \le 3$, but $xyz \to 0$, so it has no minimum. i.e. $xyz \in (0, 1]$.
Similarly, $(x+y+z)^2 \ge 3(xy+yz+zx) \ge 1 \implies x+y+z \ge 1$. As $x=y=z=\frac13$ gives equality, this is the minimum.
As $z \to 0$, we have $xy+yz+zx \to xy$ and further if we keep $x = \dfrac1y$ while $y \to 0$, we can have $xy+yz+zx \to 1$ while $x+y+z \to \infty$. Thus there is no maximum for $x+y+z$, i.e. $x+y+z \in [1, \infty)$.
