Given $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ defined as $$f(x) = x^2 + 3y^2 +2z^2 - 2xy + 2xz$$
I am trying to show $f$ attains a global minimum at the origin without using calculus. I was thinking of completing the square for a start, but I am looking for some creative ideas to employ this technique.
\begin{align*} f(x, y, z) & = (x - y + z)^2 + 2y^2 + z^2 + 2yz \\ & = (x - y + z)^2 + (y + z)^2 + y^2 \\ & \geq 0 \\ \end{align*}
The last equality holds if and only if \begin{equation} \left\{\begin{array}{r} x - y + z = 0 \\ y + z = 0 \\ y = 0 \end{array} \right. \end{equation} that is, if and only if $x = y = z = 0$. Therefore the global minimum is attained at $(0, 0, 0)$.
If you are not all that familiar with working with inequalities, you can use an approach of looking at the behavior of the function on the coordinate planes through the origin. (Picturing the function otherwise isn't all that convenient since it is a function of three variables, which would call for a four-dimensional graph to show the function values.)
In the $ \ xy-$ plane , the function is $$ \ f(x, \ y, \ 0) \ \ = \ \ x^2 \ + \ 3y^2 \ - \ 2xy \ = \ (x^2 \ - \ 2xy \ + \ y^2) \ + \ 2y^2 \ \ = \ \ ( x - y )^2 \ + \ 2y^2 \ \ . $$
The first term equals zero on the line $ \ x \ = \ y \ $ and increases with distance from the line to either side. But this function "slice" also increases with increasing absolute value of $ \ y \ $ . So the minimum value in this coordinate plane is $ \ f \ = \ 0 \ $ at $ \ x \ = \ 0 \ , \ y \ = \ 0 \ $ .
In the $ \ xz-$ plane , the function is $$ \ f(x, \ 0, \ z) \ \ = \ \ x^2 \ + \ 2z^2 \ + \ 2xz \ = \ (x^2 \ + \ 2xz \ + \ z^2) \ + \ z^2 \ \ = \ \ ( x + z )^2 \ + \ z^2 \ \ , $$
the first term of which is equal to zero on the line $ \ x \ = \ -z \ $ , but again increases with increasing distance on either side of this line; the function "slice" also increases with increasing absolute value of $ \ z \ $ . So the function in this plane is minimal ( $ \ f \ = \ 0 \ $ ) at $ \ x \ = \ 0 \ , \ z \ = \ 0 \ $ .
Finally, in the $ \ yz-$ plane , the function is $$ \ f(0, \ y, \ z) \ \ = \ \ 3y^2 \ + \ 2z^2 \ \ , $$
which is positive everywhere except at $ \ y \ = \ 0 \ , \ z \ = \ 0 \ $ .
Thus, the function is zero at the origin and positive elsewhere in $ \ \mathbb{R}^3 \ $ .
