Question: Let $a, b \in \mathbb{R}^2$ be two points in the plane and let $\Pi$ be their perpendicular bisector (see left figure). Let $c \in \Pi$ be any point and consider the triangle $\triangle abc$. Suppose that there is another point $p \in \operatorname{Int}{\triangle abc}$. It's easy to see that $d(c, p) < d(c, a) = d(c, b)$. Is the same true of a geodesic triangle on a surface $\Sigma \subset \mathbb{R}^3$? In this case we consider the intrinsic metric $d\_{\Sigma}$ defined by shortest paths on the surface $\Sigma$. We then define the 'perpendicular bisector' as the set of all points equidistant to $a, b \in \Sigma$ with respect to the metric $d\_{\Sigma}$. (See right figure.)

[![enter image description here](https://i.stack.imgur.com/Ilr8u.png)](https://i.stack.imgur.com/Ilr8u.png) [![enter image description here](https://i.stack.imgur.com/YdLb6.png)](https://i.stack.imgur.com/YdLb6.png)

**Edit:** I don't think this is true since the height of the surface could increase rapidly within the interior of the triangle. A side view, looking down in the direction of $\vec{ab}$, of what I'm talking about might looking like the following figure.

[![enter image description here](https://i.stack.imgur.com/l1kjR.png)](https://i.stack.imgur.com/l1kjR.png)

In which case, can I prove the result if I assume a bound on the principal curvatures in the interior of the triangle?

Answer: 
This is a straightforward calculation: Expanding the fraction by $\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}$ gives
\begin{align\*}
 &\, \frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}} \\
=&\, \frac{(\sqrt{n^3-3}-\sqrt{n^3+2n^2+3)}(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})}{\sqrt{n+2} (\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})} \\
=&\, \frac{(n^3-3)-(n^3+2n^2+3)}{\sqrt{(n+2)(n^3-3)} + \sqrt{(n+2)(n^3+2n^2+3)}} \\
=&\, \frac{-2n^2-6}{\sqrt{n^4+2n^3-3n-6} + \sqrt{n^4+4n^3+4n^2+3n+6}}.
\end{align\*}
Dividing numerator and denumerator by $n^2$ gves
\begin{align\*}
 &\, \frac{-2n^2-6}{\sqrt{n^4+2n^3-3n-6} + \sqrt{n^4+4n^3+4n^2+3n+6}} \\
 =&\, \frac{-2-\frac{6}{n^2}}{\sqrt{1+\frac{2}{n}-\frac{3}{n^3}-\frac{6}{n^4}} + \sqrt{1+\frac{4}{n}+\frac{4}{n^2}+\frac{3}{n^3}+\frac{6}{n^4}}}.
\end{align\*}
Taking the limit $n \to \infty$ results in $-2/2 = -1$.
When you are at the stage
$$
\lim\_{n\to\infty}\frac{-2n^2-6}{
 \sqrt{n+2}\,(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})
}
$$
pull $n^2$ from the numerator, $n$ from $n+2$ and $n^3$ from the other two radicals, so you have
$$
\lim\_{n\to\infty}\frac{n^2\left(-2-\dfrac{6}{n^2}\right)}{
 \sqrt{n}\,\sqrt{1+\dfrac{2}{n}}\,
 \sqrt{n^3}\left(
 \sqrt{1-\dfrac{3}{n^3}}+\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^3}}
 \right)
}
$$
Cancel $n^2$ with $\sqrt{n}\,\sqrt{n^3}$ and you're done.