Find the range of values of $x+y+z$

Question

If $x,y,z$ are positive reals satisfying $\frac{1}{3} \le xy+yz+ zx \le 3$, then find the range of values of $x+y+z$

This is from BMO 1993. I have applied AM GM and CS in every possible way my brain can think of. I do not think my working will be of any help to anyone, so I am not showing. Any kind of hint to full answer will be appreciated.

Answer 1

For the lower bound use the fact that $(x+y+z)^2 \geq 3(xy+xz+yz) \geq 3$ so $x+y+z \geq 1$. The equality is satifised for $ x=y=z=\frac{1}{3}$

To prove there is no upper bound pick $x=n, y=\frac{1}{n}, z=\frac{1}{n} $ for some natural number $n$. Then the conditions are satisfied.

Answer 2

We have \begin{eqnarray*} (x-y)^2+(y-z)^2+(z-x)^2 \geq 0. \\ \end{eqnarray*} Now add $3(xy+yz+zx)$ and complete the square \begin{eqnarray*} (x+y+z)^2 \geq 3(xy+yz+zx) \geq 1. \\ \end{eqnarray*} So $1$ is a lower bound.