I do not know where to start this challenge, help please

Question

If $x$, $y$ and $z$ are positive numbers such that $1\leq xy+yz+zx\leq3$ which is the set of values ​​of $xyz$? And $x + y + z$?$$$$Knowing that $x, y$ and $z$ $\in\mathbb{R^*}$

Answer 1

Hint: For the product use $$\sqrt[3]{(xyz)^2} \le \dfrac{xy+yz+zx}{3}$$ and $x\cdot 1\cdot 1 \to 0,\ x \to 0$.
For the sum use $$2(x+y+z)^2 = 2x^2+2y^2+2z^2 +4(xy+yz+zx) = x^2+y^2+x^2+z^2+y^2+z^2 +4(xy+yz+zx) \ge 6(xy+yz+zx)$$ and $x+\frac1x+\frac1x \to +\infty,\ x \to +\infty$.

So $0 < xyz \le 1$ and $\sqrt{3} \le x+y+z<+\infty$.