Determine $x, y, z$ so that they satisfy a system

Question

Determine $x, y, z$ so that they satisfy the following system:

• $0 < x \le y \le z$
• $x + y + z = 7$
• $xyz = 9$

I'd appreciate someone writing a full procedure for determining $x, y, z$.

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Correct solution is:

$$x = [1, 3/2],~ y = [3/2, 3],\ z = [3, 4]$$

Answer 1

Fix $x$. Since $y+z=7-x$ and $yz=9/x$, Vieta’s formulas imply that $y$ and $z$ are the roots of the equation $t^2+(x-7)t+9/x=0$. Since $y\le z$, we have $y=\frac 12\left(7-x-\sqrt{D}\right)$ and $z=\frac 12\left(7-x+\sqrt{D}\right)$, where $D=(7-x)^2-\frac {36}x=(x-1)(x-4)(x-9)/x\ge 0$. Now check the range for $x$ such that $y\ge x$. It yields a inequality $7-\sqrt{D} \ge 3x$. That is $0<x\le 7/3$, $0\le D\le 49$ and $(7-3x)^2\ge (7-x)^2-36/x$. The routine calculation shows that this holds iff $x\in [1, 3/2]$. Since $\frac{\partial D}{\partial x}=14-2x+\frac{36}{x^2}\ge 0$ we see that when $x$ increases then $y$ decreases. If $x=1$ then $y=3$, if $x=3/2$ then $y=3/2$, so the range of $y$ is $[3/2,3]$. To find the range for $z$ we fix it. Since $x+y=7-z$ and $xy=9/z$, Vieta’s formulas imply that $x$ and $y$ are the roots of the equation $t^2+(z-7)t+9/z=0$. Since $x\le y$, we have $x=\frac 12\left(7-z-\sqrt{D}\right)$ and $y=\frac 12\left(7-z+\sqrt{D}\right)$, where $D=(7-z)^2-\frac {36}z=(z-1)(z-4)(z-9)/z\ge 0$ (which is equivalent to $1\le z\le 4$). Now check the range for $z$ such that $y\le z$. It yields a inequality $7+\sqrt{D} \le 3z$. That is $7/3\le z\le 4$ and $(3z-7)^2\ge (7-z)^2-36/z$. Since the last inequality is equivalent to $(z+1)(z-3)(2z-3)\ge 0$ , we see that the range of $z$ is $[3,4]$.