Is this function on $\mathbb{R}^{2}$ always positive? Consider the following function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2} $:
$$f(x):=\alpha\left( |x-y|^{\alpha-2}(x-y)-|x-z|^{\alpha-2}(x-z) \right)$$
Where $\alpha >0,\alpha \neq 1$. I'm trying to show that
$|f|>0$ as long as $y\neq z$, $x\neq y$, $x\neq z$.
We have
$$|f (x)|^2=$$
$$|x-y|^{2\alpha-2}
-2|x-y|^{\alpha-2}|x-z|^{\alpha-2}\langle x-y,x-z\rangle +|x-z|^{2\alpha-2}$$
To simplify, denote $a=x-y$ and $b=x-z$. Then
$$|f (x)|^2=|a|^{2\alpha-2}
-2|a|^{\alpha-2}|b|^{\alpha-2}\langle a,b\rangle +|b|^{2\alpha-2}$$
To minimize this we need maximize $\langle a,b\rangle$
if its positive (and minimize $\langle a,b\rangle$ if it is negative).
From Cauchy's inequality
$$|\langle a,b\rangle|\leq |a||b|$$
and equality happens if and only if
$$b=c a,$$ where $c$ is a scalar. In this case
$$\langle a,b\rangle=c|a|^2.$$
The problem then reduces to minimize for $c>0$ the function
$$c\mapsto |f (x)|^2=|a|^{2\alpha-2}
-2c^{\alpha-1}|a|^{2\alpha-2} +c^{2\alpha-2}|a|^{2\alpha-2}
=|a|^{2\alpha-2}(1-2c^{\alpha-1} +c^{2\alpha-2} )
=|a|^{2\alpha-2}(1-c^{\alpha-1})^2$$
which is strictly positive if and only if $a\neq 0 \Longleftrightarrow x\neq y$ and $c\neq 1 \Longleftrightarrow a\neq b\Longleftrightarrow x-y\neq x-z\Longleftrightarrow y\neq z$.
Is my argument correct?
 A: Your argument looks good. For fun and posterity, here's a geometric/physical argument:
Let $\alpha \neq 1$ be an arbitrary positive real. For a fixed vector $y$, the vector field
$$
F_{y}(x) = \alpha |x - y|^{\alpha-2}(x - y)
$$
is non-vanishing away from $y$, points radially away from $y$, and its magnitude is strictly monotone along each ray (decreasing if $0 < \alpha < 1$ and increasing if $1 < \alpha$).
By definition, $f(x) = F_{y}(x) - F_{z}(x)$. Let's call the terms on the right ($F_{y}$ and $-F_{z}$) the summands.
If $x$ does not lie on the line $\overline{yz}$, then $f(x) \neq 0$ because the summands are non-proportional, hence linearly independent.
To handle the remaining cases, partition the line $\overline{yz}$ into

*

*The open ray away from $y$ and not containing $z$;

*The open ray away from $z$ and not containing $y$;

*The open segment from $y$ to $z$.

On each unbounded ray, the summands are oppositely proportional, but have different magnitude because $x$ is farther from one point ($y$ or $z$) than from the other, so $f(x) \neq 0$.
On the segment, the summands are positively proportional (away from $y$ is towards $z$), so their sum is non-zero.
In all cases, the field $f$ is non-vanishing away from $y$ and $z$, so its magnitude is positive: $|f| > 0$.
