Find the minimum value !

The equation $x^2+y^2+z^2 = 4$ is even, where both positive and negative values of $x$ , $y$ , and $z$ satisfy the equation. Therefore, if we find the maximum value of $12x+9y+8z$ , which is positive, the minimum will have the same absolute value of the maximum but with a negative sign.

Considering the positive real value of $x$ , $y$ , and $z$ using Cauchy-Schwarz inequality :

$\begin{aligned} (12x+9y+8z)^2 & \le (12^2+9^2+8^2)(x^2+y^2+z^2) & \small \color{#3D99F6} \text{Equality occurs when } \\ & = 289 \times 4 = 1156 & \small \color{#3D99F6} x = \frac {24}{17}, y = \frac {18}{17}, z = \frac {16}{17} \end{aligned}$

$\implies 12x+9y+8z \le 34$ for positive reals $x$ , $y$ , and $z$ .

Since $\max(12x+9y+8z) = 34$ , $\min(12x+9y+8z) = \boxed{-34}$ .

Use the dot product : $12x+9y+8z=(12,9,8) \cdot (x,y,z)=||(12,9,8)||\cos\theta ||(x,y,z)|| =17\times 2\cos \theta \geq \boxed{-34}$ ; equality is attained when $\theta = \pi$ .

The first equation is of a sphere in the 3d plane with radius 2. The second equation is of a plane $12x+9y+8z=c$ , where we have to find the minimum c. Notice that the maximum value of c = minimum value in magnitude.

Max value of c will be obtained when this plane is just touching this sphere, i.e distance of plane from sphere centre (origin) = radius

$|\frac {0+0+0-c}{\sqrt {12^2 +9^2+8^2}}|=2$

Solve this to get $c=±34$