minimum function on a set

Question

How to find the minimum $x+y+z$ on the following set $ (x-2)^{2}+(y-2)^{2}+(z-2)^{2} \le 1 ? $ is it $1$? I am just guessing some of the points $(1,0,0), (0,1,0), (0,0,1)$. I have no clue how to solve these examples.

Answer 1

Let $f (x,y,z)=x+y+z $

$(x-2)^2+(y-2)^2+(z-2)^2 \leq 1$ spans a sphere which has a center C(2,2,2)

$f_x (x,y,z)=f_y (x,y,z)=f_z (x,y,z)=1 \gt 0$ that means f is increasing so

x < a, y <b , z<c $\Rightarrow f (x,y,z) < f (a,b,c) $

If there exists a point O which makes f minimum it must be close to smaller values as much as possible so the point P is on the $(x-2)^2+(y-2)^2+(z-2)^2=1$

That is important because we can use Lagrange's method now

Let $g (x,y,z)=(x-2)^2+(y-2)^2+(z-2)^2 - 1=0$ then there exists a scalar k which makes $\nabla f (x,y,z)= k \nabla g(x,y,z) $ for point P (x,y,z) $1=(2x-2)k$

$1=(2y-2)k$

$1=(2z-2)k$

$2x-2=2y-2=2z-2 \Rightarrow x=y=z $

$3(x-2)^2=1$ $ \Rightarrow x=2-\frac {1}{\sqrt (3)} $

$f(x,y,z)=6-\sqrt (3) $

Answer 2

For $x=y=z=2-\frac{1}{\sqrt3}$ we obtain the value $6-\sqrt3$.

We'll prove that it's a minimal value.

Indeed, $$\sum_{cyc}\left(x-2+\frac{1}{\sqrt{3}}\right)=$$ $$=\sum_{cyc}\left(x-2+\frac{1}{\sqrt{3}}+\frac{\sqrt3}{2}\left(x-2)^2-\frac{1}{3}\right)\right)+\frac{\sqrt3}{2}\left(1-\sum_{cyc}(x-2)^2\right)=$$ $$=\frac{\sqrt3}{2}\sum_{cyc}\left(x-2+\frac{1}{\sqrt{3}}\right)^2+\frac{\sqrt3}{2}\left(1-\sum_{cyc}(x-2)^2\right)\geq0$$