If $x,y,z$ are positive reals satisfying $\frac{1}{3} \le xy+yz+ zx \le 3$, then find the range of values for $xyz$
This is actually the first part of the problem. The next part is to find the range of values for $x+y+z$, which I posted previously and got perfect answers. I do not think my working will be of any help to the readers. So any help from hints to full answers will be appreciated. Now my problem is to find the range of values for $xyz$ and not $x+y+z$ so please fo not close this and provide me a link to my previously asked question!
For the lower bound, $0<xyz$ since they are positive, but $xyz$ can be arbitrarily close to $0$ - for any $\delta>0$ take $x=\delta,y=z=1$.
For the upper bound, AM/GM gives $\frac{xy+yz+zx}{3}\geq\sqrt[3]{x^2y^2z^2}$, and since $xy+yz+zx\leq 3$ we get $xyz\leq 1$. Equality is possible if $x=y=z=1$.
So the range is $0<xyz\leq 1$.