For $p\geq 1$ let $D_p = \{(x,y)\in \mathbb R^2: y> |x|^{1/p}\}$. I would like some help to prove or disprove the following statement: There exists $\epsilon>0$ such that for all $0< r< 1$, the containment $\{(x,y)\in \mathbb R^2: x^2 + (y - r)^2 < \epsilon^2 r^{2p}\}\subset D_p$ holds.
It is easy to see that the statement is true in the case $p = 1$ (for example, $\epsilon = \frac{1}{2\sqrt 2}$ works with room to spare) because one can easily compute the distance from the point $(0, r)$ to the boundary of $D_1$. The case $p>1$ is giving me trouble. In fact, I'd even be happy if I could gain some insight for the case $p = 2$.
The case $p=1$ is simple, and needs special treatment. In the following I assume $p>1$ and reflect the figure on the line $x=y$. We therefore consider the curve $$\gamma:\quad t\mapsto{\bf z}(t):=(t,t^p)\qquad(t\geq0)$$ with tangent vector ${\bf z}'(t)=(1,pt^{p-1})$. Fix a $t>0$. The downwards normal $\nu$ at ${\bf z}(t)$ is given by $$\nu:\quad s\mapsto {\bf n}(s)=(t,t^p)+s(p t^{p-1},-1)\qquad(s\geq0)\ .$$ This normal intersects the $x$-axis when $s=t^p$ at the point ${\bf r}(t)=\bigl(r(t),0\bigr)$, whereby $$r(t)=t+p t^{2p-1}\ .\tag{1}$$ As $\gamma$ is convex the function $t\mapsto r(t)$ is a monotonically increasing function of $t$. The distance $\rho(t):=|{\bf z}(t)-{\bf r}(t)|$ computes to $$\rho(t)=t^p\sqrt{1+p^2 t^{2p-2}}\ .\tag{2}$$ For your problem we have to express $t$ as a function of $r$, using $(1)$. One obtains $$t=r-\bigl(p+o(1)\bigr)r^{2p-1}\qquad(r\to0+)\ .$$ Now you have to plug this into $(2)$ and obtain, after some calculations, $$\rho=r^p\bigl(1+ (c+o(1)) r^\alpha\bigr)\qquad(r\to0+)$$ whereby I leave the value of the $c$ and the $\alpha>0$ to you.