It is shown in the post
Is this function on $\mathbb{R}^{2}$ always positive?
that the modulus $|f|$ of the function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2} $ given by :
$$f(x):=\alpha\left( |x-y|^{\alpha-2}(x-y)-|x-z|^{\alpha-2}(x-z) \right)$$ where $\alpha >0,\alpha \neq 1$ is strictly positive when $y\neq z$.
I am trying to get a uniform lower bound of $|f|$ given that $|y-z|> 1$. By uniform I mean in uniform in $x$.
Any hints ?
Some thoughts: Let $t=x-y$. Denote $\eta=y-z$ so that $x-z=t+\eta$. Then it suffices to find a (uniform in $t$) lower bound for
$$g(t):=\left| |t|^{\alpha-2}t-|t+\eta|^{\alpha-2}(t+\eta) \right|$$
I had earlier attempted the bound $$\left| |t|^{\alpha-2}t-|t+\eta|^{\alpha-2}(t+\eta) \right|\geq \left| |t|^{\alpha-1}-|t+\eta|^{\alpha-1} \right|$$ but the latter is not bounded away from zero because (as @Calvin Khor pointed out) it attains zero at $t=-\eta/2$.
An update: When $x$ lies on the perpendicular bisector $L$ of $\overline{yz}$ we have $|f(x)|\geq \frac{1}{2^{\alpha-2}}|y-z|^{\alpha-1}$. Why? For every $x$ on the bisector we have $|x-y|=|x-z|$. Therefore, for any $x\in L$, $|f(x)|=|x-y|^{\alpha-2}|x-y-(x-z)|=|x-y|^{\alpha-2}|y-z|$. If $x$ is the midpoint of $\overline{yz}$ then $|x-y|=\frac{1}{2}|y-z|$. If $x \in L$ then $|x-y|>\frac{1}{2}|y-z|$,
tl; dr: The magnitude of the vector field $f$ is bounded below if and only if $2 \leq \alpha$.
As in the linked question, let's fix $\alpha > 0$ and write $$ F_{y}(x) = \alpha|x - y|^{\alpha-2}(x - y). $$ The vector field $F_{y}$ points radially away from $y$, and has magnitude $|F_{y}(x)| = \alpha|x - y|^{\alpha-1}$ that is constant on circles centered at $y$. The magnitude is monotone in the distance $|x - y|$, decreasing if $0 < \alpha < 1$ and increasing if $1 < \alpha$.
Since $f(x) = F_{y}(x) - F_{z}(x)$, the summands $F_{y}(x)$ and $-F_{z}(x)$ have equal magnitude on the perpendicular bisector of the segment $\overline{yz}$, where $|x - y| = |x - z|$.
Since $|f| > 0$ everywhere $f$ is defined (away from $y$ and $z$ for all $\alpha$, and everywhere in the plane if $1 < \alpha$) and the magnitude is continuous in $x$, the extreme value theorem guarantees there is a positive lower bound on $|f|$ if and only if there is a positive lower bound "at infinity". (If the field is undefined at $y$ and $z$, the magnitude is unbounded at these points.)
To proceed further, let's put $2c = |y - z|$ and choose Cartesian coordinates $(x_{1}, x_{2})$ so that $y = (-c, 0)$ and $z = (c, 0)$, so that $$ f(x_{1}, x_{2}) = \alpha[|(x_{1} + c, x_{2})|^{\alpha-2}(x_{1} + c, x_{2}) - |(x_{1} - c, x_{2})|^{\alpha-2}(x_{1} - c, x_{2})]. $$ Introducing the expressions \begin{align*} A_{m} &= |(x_{1} + c, x_{2})|^{\alpha-2} - |(x_{1} - c, x_{2})|^{\alpha-2}, \\ A_{p} &= |(x_{1} + c, x_{2})|^{\alpha-2} + |(x_{1} - c, x_{2})|^{\alpha-2}, \end{align*} we have $$ f(x) = \alpha(A_{m} x_{1} + A_{p} c, A_{m} x_{2}). $$
Along the perpendicular bisector, i.e., the line $x_{1} = 0$, we have $A_{m} = 0$ and $A_{p} = 2|c^{2} + x_{2}^{2}|^{(\alpha-2)/2}$, so $$ f(0, x_{2}) = 2c\alpha|c^{2} + x_{2}^{2}|^{(\alpha-2)/2}(1, 0). $$ If $0 < \alpha < 2$, the magnitude $|f(0, x_{2})| = 2c\alpha|c^{2} + x_{2}^{2}|^{(\alpha-2)/2}$ has no positive lower bound for real $x_{2}$, so a fortiori $|f(x)|$ has no positive lower bound in the plane.
If $\alpha = 2$, then $$ f(x_{1}, x_{2}) = \alpha[(x_{1} + c, x_{2}) - (x_{1} - c, x_{2})] = \alpha(2c, 0) $$ is constant as a vector field.
If $2 < \alpha$, the magnitude of each summand is convex. Qualitatively, we expect a positive lower bound on $|f|$ because:
[Added: This outline is not how the estimate below proceeds, it's just offered as evidence that we should look for a positive lower bound if $2 < \alpha$. The estimate below is corrected and expanded; initially I expanded the power functions to first order, but the second order term is needed to obtain the stated accuracy.]
Again, we need only establish a positive lower bound for sufficiently large $|x|$. To this end, let's write, for $0 < p = \alpha - 2$ real and $|x|\gg 1$, \begin{align*} |(x_{1} \pm c, x_{2})|^{p} &= ((x_{1} \pm c)^{2} + x_{2}^{2})^{p/2} \\ &= (|x|^{2} \pm 2cx_{1} + c^{2})^{p/2} \\ &= |x|^{p}\, \Bigl|1 + \frac{\pm2cx_{1}+c^{2}}{|x|^{2}}\Bigr|^{p/2}. \end{align*} Setting \begin{align*} u &= \frac{\pm2cx_{1} + c^{2}}{|x|^{2}} = O\Bigl(\frac{1}{|x|}\Bigr), \\ u^{2} &= \frac{4c^{2}x_{1}^{2} \pm4c^{3}x_{1} + c^{4}}{|x|^{4}} = \frac{4c^{2}x_{1}^{2}}{|x|^{4}} + O\Bigl(\frac{1}{|x|^{3}}\Bigr) \end{align*} in the second-order Taylor approximation $$ (1 + u)^{p/2} = 1 + \tfrac{1}{2}pu + \tfrac{1}{8}p(p - 2)u^{2} + O(u^{3}), $$ we have $$ |(x_{1} \pm c, x_{2})|^{p} = |x|^{p}\, \Bigl[1 + \frac{p}{2}\, \frac{\pm2cx_{1}+c^{2}}{|x|^{2}} + \frac{p(p - 2)}{8}\, \frac{4c^{2} x_{1}^{2}}{|x|^{4}} + O\Bigl(\frac{1}{|x|^{3}}\Bigr)\Bigr]. $$ Substituting, and using the notation above, \begin{align*} A_{m} = |(x_{1} + c, x_{2})|^{p} - |(x_{1} - c, x_{2})|^{p} &= |x|^{p}\, \Bigl[\frac{2pcx_{1}}{|x|^{2}} + O\Bigl(\frac{1}{|x|^{3}}\Bigr)\Bigr], \\ A_{p} = |(x_{1} + c, x_{2})|^{p} + |(x_{1} - c, x_{2})|^{p} &= |x|^{p}\, \Bigl[2 + \frac{pc^{2}}{|x|^{2}}\Bigl(1 + \frac{(p - 2)x_{1}^{2}}{|x|^{2}}\Bigr) + O\Bigl(\frac{1}{|x|^{3}}\Bigr)\Bigr] \\ &= |x|^{p}\, \Bigl[2 + O\Bigl(\frac{1}{|x|^{2}}\Bigr)\Bigr]. \end{align*} Thus, with $p = \alpha - 2 > 0$, \begin{align*} f(x) &= \alpha(A_{m} x_{1} + A_{p} c, A_{m} x_{2}) \\ &= \alpha|x|^{\alpha-2} \biggl(\frac{2pc x_{1}}{|x|^{2}} x_{1} + 2c + O\Bigr(\frac{1}{|x|^{2}}\Bigr), \frac{2pc x_{1}}{|x|^{2}} x_{2} + O\Bigr(\frac{1}{|x|^{2}}\Bigr)\biggr). \end{align*} Modulo lower-order terms (i.e., in the limit as $|x| \to \infty$), the vector in parentheses has first component at least $2c$, so its magnitude is at least $2c$. Consequently $$ |f(x)| > c\alpha|x|^{\alpha-2} $$ for sufficiently large $|x|$.
[Added: Here is an extreme value theorem argument to show $|f|$ has a positive lower bound. The preceding estimate shows there exists an $R > 0$ such that
By the extreme value theorem, the continuous, positive function $|f(x)|$ has a positive absolute minimum in the closed disk of radius $R$ about $0$. That is, there exists an $x_{0}$ with $|x_{0}| \leq R$ such that $|f(x)| \geq |f(x_{0})| > 0$ for all $x$ with $|x| \leq R$. Particularly, $|f(0)| \geq |f(x_{0})|$.
If instead $|x| > R$, then by the choice of $R$ we have $$ |f(x)| > c\alpha|x|^{\alpha-2} > c\alpha R^{\alpha-2} > |f(0)| \geq |f(x_{0})|. $$ We have therefore shown $|f(x)| \geq |f(x_{0})| > 0$ for all $x$ in the plane, i.e., that $|f(x)|$ has a positive lower bound.]