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

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.

• Please, anyone's any help will be appreciated!😢😢 – ami_ba Oct 20 '17 at 19:22

## 2 Answers

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.

• Perfect! Thanks. – ami_ba Oct 20 '17 at 19:32
• Can you also tell the range for $xyz$? – ami_ba Oct 20 '17 at 19:36
• The upper bound is $1/27$ and the lower bound is $0$. – velut luna Oct 20 '17 at 19:41
• So what did you apply? – ami_ba Oct 21 '17 at 5:55
• The upper bound for xyz follows from AM-GM applied for xy+yz+zx. You then get xyz <=1. For the lower bound apply the same example as in the answer. – C.Niculescu Oct 21 '17 at 9:38

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.