Dirty Minima

We have ${ x }^{ 2 }+2{ y }^{ 2 }+{ z }^{ 2 }+2\left( xy+z-2y \right)$

Which can be rearranged to get ${ x }^{ 2 }+{ y }^{ 2 }+2xy+4+{ y }^{ 2 }-4y+{ z }^{ 2 }+1+2z-5$

Using identity over these terms ${ \left( x+y \right) }^{ 2 }+{ \left( 2-y \right) }^{ 2 }+{ \left( z+1 \right) }^{ 2 }-5$

Since each of these terms are non-negative, we need to equate each of them to 0 to obtain minima (if possible).

And this is possible as we can see that $y=2,\quad x=-y,\quad z=-1$ can be obtained simultaneously.

This brings the minima to ${ 0 }^{ 2 }+{ 0 }^{ 2 }+{ 0 }^{ 2 }-5=\boxed{-5}$

$s = x^2 + 2y^2 + z^2 + 2(xy + z - 2y)$

First:

$\frac{\partial s}{\partial x} = 2x + 2y$

And equal to zero, so:

$x = -y$

Substitute x:

$s = 3y^2 + z^2 + 2(-y^2 + z - 2y)$

Second:

$\frac{\partial s}{\partial z} = 2z + 2$

And equals to zero, so:

$z = -1$

Substitute z:

$s = y^2 - 4y - 1$

Third:

$\frac{\partial s}{\partial y} = 2y - 4$

And equals to zero, so:

$y = 2$

Substitute y:

$s =-5$