For ease of notation, I'll make a quick relabeling of the problem. Suppose $D$ is a point in the interior of equilateral triangle $ABC$, and $E$ is a point inside $\triangle DBC$. We show that \[\frac{[DBC]}{P(DBC)^2}>\frac{[EBC]}{P(EBC)^2}.\] Let $\angle B=2x, \angle C=2y, \angle D=2z$. Then we have $[DBC]=x\cdot\frac{r}{2}\cdot P(DBC).$ We also have $BC=r\cot{x}+r\cot{y}$ and similarly for the other sides. Thus, $P(DBC)=2r\sum\cot{x},$ and it follows that $\frac{[DBC]}{P(DBC)}=\left(4\sum\cot{x}\right)^{-1}.$ Let this value be $T$. We can set $z=90^{\circ}-x-y$, to get \[\frac{1}{4}T=\cot{x}+\cot{y}+\frac{\cot{x}+\cot{y}}{\cot{x}\cot{y}-1}.\] Since $\triangle EBC$ has both $x$ and $y$ smaller than in $\triangle DBC$, and since $\cot{x}$ is decreasing on the range $0$ to $30^{\circ}$, so it suffices to show that \[f(u,v)=u+v+\frac{u+v}{uv-1}\] is an increasing function in $u$, where $u=\cot{x}, v=\cot{y}.$ Since $f$ is symmetric it is also an increasing function in $v$. We have $x,y<30^{\circ}$, so we want to show that $f$ is increasing for $u,v>\sqrt{3}.$ We have \begin{align*}\frac{\partial f}{\partial u}&=\frac{v(u^2v-2u-v)}{(uv-1)^2}\\ &=\frac{\cot{y}\left[\cot^2{x}\cot{y}-2\cot{x}-\cot{y}\right]}{(\cot{x}\cot{y}-1)^2}\\ &=\cos{y}\cos(2x+y)\sec^2(x+y),\end{align*} after a boxload of algebra. This is clearly positive for the values of $x$ and $y$ we are working with, so $f$ is increasing in both $u$ and $v$, so we are done.