Range of a Three Variable Fun-ction !!

For the sake of variety, let me submit a geometrical solution; I like to see what I am doing.

Using the dot product, we can write $f(x,y,z)=\frac{(x,y,z)\cdot(y,z,x)}{||(x,y,z)||^2}=\cos(\theta)$ , where $\theta$ is the angle between the vectors $(x,y,z)$ and $(y,z,x)$ . Now $(y,z,x)$ is obtained by rotating $(x,y,z)$ through $2\pi/3$ about the line spanned by $(1,1,1)$ , a permutation of the axes. Thus $0\leq\theta\leq2\pi/3$ and $[a,b]=[\cos(2\pi/3),\cos(0)]=[-1/2,1].$ Finally, $a+b=\boxed{0.5}$ .

$We\quad know\quad that:\\ { \left( x+y+z \right) }^{ 2 }=\left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }+2xy+2yz+2xz \right) \quad \ge \quad 0\\ \Rightarrow \quad \frac { xy+yz+xz }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \ge \quad -\frac { 1 }{ 2 } \quad ...........(i)\\ Now:\\ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }-xy-yz-xz=\cfrac { 1 }{ 2 } \left( { \left( x-y \right) }^{ 2 }{ +\left( y-z \right) }^{ 2 }+{ \left( x-z \right) }^{ 2 } \right) \quad \ge \quad 0\\ \Rightarrow \frac { xy+yz+xz }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \le \quad 1\quad \quad \quad \quad .............(ii)\\ By\quad (i)\quad and\quad (ii)\quad we\quad can\quad say\quad that\quad range\quad of\quad \\ f(x,y,z)\quad is\quad \left[ \frac { -1 }{ 2 } ,1 \right] \\ \therefore \quad \boxed { \quad a+b\quad =\quad 0.5\quad }$

SInce $(x+y+z)^2 = (x^2+y^2+z^2)+2(xy+yz+zx),$ the given function is equal to $f(x,y,z) = \frac12\left(\frac{(x+y+z)^2}{x^2+y+2+z^2}-1\right).$ Interpreting $R^2 = x^2+y^2+z^2$ as the square radius of a sphere centered at zero, and $x+y+z$ as the dot product of a vector lying on that sphere with $1, 1, 1$ .

Maximum value: Choose vector on the sphere parallel to $1, 1, 1$ . The coordinates are $x=y=z=\pm\tfrac13\sqrt3R$ and we find $|x + y + z| = \sqrt 3R.$

Minimum value: Choose vectors on the sphere perpendicular to $1, 1, 1$ . There are infinitely many such vectors (lying on a ring). They satisfy $|x + y + z| = 0$ .

Substituting the minimum and maximum values of $|x+y+z|$ into the equation, we find that $f$ lies between $\tfrac12((\sqrt3)^2-1) = 1$ (max) and $\tfrac12(0^2-1) = -\tfrac12$ (min).

Sorry but i think that this question is overrated. It should be of level 3