Min + Max?

Let $f=xyz$ , then

$\begin{cases} x+y+z=3 \\ xy+yz+zx=-9 \\ xyz=f \end{cases}$

so $x,y,z$ are 3 roots of the equation $t^3-3t^2-9t-f=0 \quad (1)$ (by applying Viète theorem)

Let $u=t-1$ then $(1)$ becomes $u^3-12u-f-11=0 \quad (2)$

Since $x,y,z$ are real numbers, equation $(2)$ have 3 real roots, that comes to:

$\frac { { \left( -f-11 \right) }^{ 2 } }{ 4 } +\frac { { \left( -12 \right) }^{ 3 } }{ 27 } \le 0\Longleftrightarrow -27\le f\le 5$ .

+) $f=5$ , then $x,y,z$ are 3 roots of the equation $t^3-3t^2-9t-5=0$ , which are: $(x,y,z)=(-1,-1,5),(-1,5,-1),(5,-1,-1)$

+) $f=-27$ , then $x,y,z$ are 3 roots of the equation $t^3-3t^2-9t+27=0$ , which are: $(x,y,z)=(-3,3,3),(3,-3,3),(3,3,-3)$

Hence, $m=-27$ and $M=5$ , which gives $m+M=-22$

Slightly different solution with Linkin Duck 's

Let $xyz = c$ , then

$\begin{cases} x + y + z = 3 \\ xy+yz+zx = - 9 \\ xyz = c \end{cases}$

This means that $x$ , $y$ and $z$ are roots of $u^3-3u^2-9u-c = 0$ . Let $f(u) = u^3-3u^2-9u-c$ and let us find local maximum and minimum of $f(x)$ .

$\begin{aligned} f'(x) & = 3u^2 - 6u - 9 & \small \color{#3D99F6} \text{Equating }f'(x) = 0 \\ 3u^2 - 6u - 9 & = 0 \\ u^2 - 2u - 3 & = 0 \\ (u+1)(u-3) & = 0 \\ \implies u & = -1, \ 3 \end{aligned}$

We note that when $\begin{cases} u = -1 & f''(-1) = 6(-1)-6 < 0 & \implies f(-1) = 5+c \text{ is a maximum.} \\ u = 3 & f''(-1) = 6(3)-6 > 0 & \implies f(3) = -27+c \text{ is a minimum.} \end{cases}$

For all three roots $x$ , $y$ and $z$ to be real, $-27 \le c \le 5$ , $\implies m = -27$ , $M = 5$ and $m+M = \boxed{-22}$

From $x+y+z=3$ and $xy+yz+zx=-9$ , we get $x^2+y^2+z^2=27$ , by using the identity $(x+y+z)^2=x^2+y^2+z^2-2(xy+yz+zx)$ .

From this, we write $x+y=3-z$ and $x^2+y^2=27-z^2$

By using Cauchy-Schwarz Inequality , we have that $x^2+y^2 \ge \dfrac{(x+y)^2}{2}$ $\implies$ $27-z^2 \ge \dfrac{(3-z)^2}{2}$ which gives bounds on $z$ , that is $z \in [-3,5]$ .

If we take minimum value of $z$ , that is $z=-3$ , we get $x=y=3$ , and hence $xyz=-27$

If we take maximum value of $z$ , that is $z=5$ , we get $x=y=-1$ , and hence $xyz=5$

and summing these values yields $m+M=\boxed{-22}$

---

I don't know why taking the minimum and maximum values of $z$ , gives the answer and also I don't know why $-27$ and $5$ are minimum and maximum.

I just took the extreme bounds of $z$ , and got whatever value of $xyz$ , i assumed it to be minimum and maximum value respectively.

Please tell me how to verify why this will work. ( Also, see whether this can be done without calculus)

@Chew-Seong Cheong , @Linkin Duck