Finding the minimum of $v^2 + x^2 - 2xv\gamma$ Let's say I have this expression $v^2 + x^2 - 2xv\gamma$, in which $\gamma$ is a constant. I want to find values for $v$ and $x$ to minimize the expression.
Sounds easy: just differentiate and set the result to zero. However, differentiating with respect to $v$ yields different results compared to differentiating w.r.t $x$.
$\partial/\partial v \rightarrow 2v -2x \gamma = 0 \rightarrow v = x\gamma$
$\partial/\partial x \rightarrow 2x -2v \gamma = 0 \rightarrow x = v\gamma$
Why is there a difference? Like, the first equation implies that this function hits an extremum when $v = x \gamma$. The value of the expression at that extremum is now $\gamma^2 x^2 + x^2 - 2x^2 \gamma^2 = x^2 - x^2 \gamma^2$. On the other hand, the second equation implies that this function hits an extremum when $v = x / \gamma$, and that the value at the extremum is actually $x^2/\gamma^2 - x^2$. The two expressions are plainly different (unless $\gamma = \pm 1$).
How do I interpret this discrepancy? What exactly are the two equations $v = x\gamma$ and $x = v \gamma$? What is the true value of the expression at the extremum?

We have to go over a few cases to analyze your algebraic expression.

I have included all possible cases in my answer and I want to point out that I will use only algebra.
Using the method of completing the square, we have
$$v^2 + x^2 - 2xv\gamma$$
$$x^2-x(2v\gamma)+v^2$$
$$\begin{align}(x-v\gamma)^2+v^2-v^2\gamma^2≥v^2(1-\gamma^2)\end{align}$$
$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 - 2xv\gamma]=v^2(1-\gamma^2), ~\text{at} ~ x=v\gamma}}}\end{align}$$
Then note that,
$$x^2 + v^2 - 2vx\gamma$$
$$v^2-v(2x\gamma)+x^2$$
$$\begin{align}(v-x\gamma)^2+x^2-x^2\gamma^2≥x^2(1-\gamma^2)\end{align}$$
$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 - 2xv\gamma]=x^2(1-\gamma^2),~\text{at} ~v=x\gamma. }}}\end{align}$$

Based on the $\gamma$ constant, both results can be equal:
$$v^2(1-\gamma^2)=x^2(1-\gamma^2)$$
$$\gamma =±1$$
which implies

*

*If $\gamma =1$,

$$v^2 + x^2 - 2xv\gamma=(v-x)^2≥0$$
$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 - 2xv\gamma]=0, \text{at} ~ v=x  ~ \text{or} ~x=v}}}\end{align}$$

*

*If $\gamma =-1$,

$$v^2 + x^2 + 2xv\gamma=(v+x)^2≥0$$
$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 + 2xv\gamma]=0, \text{at} ~ v=-x ~ \text{or} ~x=-v}}}\end{align}$$

If $\gamma ≠±1$, then

*

*$|\gamma|>1$ and $x\to\infty$, then

$$\text{min}[v^2 + x^2 - 2xv\gamma] \to\ -\infty$$

*

*$|\gamma|>1$ and $v\to\infty$, then

$$\text{min}[v^2 + x^2 - 2xv\gamma] \to\ -\infty$$
This means, if $ |\gamma|>1$ , we  get
$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 - 2xv\gamma] = -\infty.}}}\end{align}$$


*

*If $|\gamma|<1$, then minimum value of your expression occurs at $x=0$ or $v=0$, which follows

$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 - 2xv\gamma]=x^2(1-\gamma^2)=0, ~\text{at} ~x=0, |\gamma|<1 }}}\end{align}$$
$$\begin{align}\color {gold}{\boxed {\color{black}{\text{min}[v^2 + x^2 - 2xv\gamma]=v^2(1-\gamma^2)=0, ~\text{at} ~v=0, |\gamma|<1.}}}\end{align}$$
