Given $\triangle ABC$, find point $X$ that minimizes $|\overline{AX}||\overline{BX}|+|\overline{BX}||\overline{CX}|+|\overline{CX}||\overline{AX}|$ Given a $\triangle$$ABC$.Find with proof the point $X$ in the plane of the triangle,for which $$|\overline{AX}||\overline{BX}|+|\overline{BX}||\overline{CX}|+|\overline{CX}||\overline{AX}|$$ is as small as possible.
 A: Let $a = |BC|$, $b = |CA|$, $c = |AB|$, $R_1 = |XA|$, $R_2 = |XB|$, $R_3 = |XC|$.  
WOLOG, we will assume $a \ge b \ge c$. We will show that
$$R_1R_2 + R_2R_3 + R_3R_1 \ge \min(ab,bc,ca) = bc \tag{*1}$$
This implies vertex $A$ is always a point that minimize $R_1R_2 + R_2R_3+R_3R_1$.
When $X \in \{ A, B, C \}$, $(*1)$ is trivially true. 
Let us consider the case $X \notin \{ A, B, C \}$ which means $R_1, R_2, R_3 > 0$.  
For any positive numbers $x, y, z$, it is well known the expression
$x R_1^2 + yR_2^2 + zR_3^2$ is minimized when and only when $X$ equals to the weighted centroid $\frac{xA + yB + zC}{x+y+z}$. By evaluating the value of this expression at that particular point, we obtain the so called "polar moment of inertia inequality" by M.S. Klamkin${}^{\color{blue}{[1]}}$,
$$(x+y+z)(xR_1^2 + yR_2^2 + zR_3^2) \ge a^2 yz + b^2 xz + c^2 xy
= (xyz)\left(\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z}\right)
$$
Using Engel's form of Cauchy Schwarz inequality, we obtain
$$
\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z}
\ge \frac{(a+b+c)^2}{x+y+z}
\quad\implies\quad
\frac{R_1^2}{yz} + \frac{R_2^2}{xz} + \frac{R_3^2}{xy} \ge
 \left(\frac{a + b + c}{x + y + z}\right)^2$$
The inequality on RHS is valid for any four distinct points $X, A, B, C$. If will continue to work if we replace $A, B, C$ by $A', B', C'$, the image of $A,B,C$ under inversion with respect to the unit circle centered at $X$.
For the new triangle $\triangle A'B'C'$, we have
$$a' = \frac{a}{R_2R_3},\; b' = \frac{b}{R_1R_3},\; c' = \frac{c}{R_1R_2},\; R_1' = \frac{1}{R_1},\; R_2' = \frac{1}{R_2},\; R_3' = \frac{1}{R_3}$$
This leads to
$$\begin{align} & \frac{(R'_1)^2}{yz} + \frac{(R'_2)^2}{xz} + \frac{(R'_3)^2}{xy} \ge
 \left(\frac{a' + b' + c'}{x + y + z}\right)^2 \tag{*2}\\
\iff &
\frac{(R_2R_3)^2}{yz} + \frac{(R_1R_3)^2}{xz} + \frac{(R_1R_2)^2}{xy}
\ge \left(\frac{aR_1 + bR_2 + cR_3}{x + y + z}\right)^2
\end{align}
$$
Chose $x : y : z = aR_1 : bR_2 : cR_3$, we obtain an inequality first proved
by T. Hayashi${}^{\color{blue}{[2][3]}}$:
$$\frac{R_2R_3}{bc} + \frac{R_1R_3}{ac} + \frac{R_1R_2}{ab} \ge 1\tag{*3}$$
Since $a \ge b \ge c$, we have $bc \le ac \le ab$ and hence
$$R_2R_3 + R_1R_3 + R_1R_2 \stackrel{*}{\ge} bc\left(\frac{R_2R_3}{bc} + \frac{R_1R_3}{ac} + \frac{R_1R_2}{ab}\right) \ge bc$$
Please note that inequality marked by $(*)$ in last line will be strict unless $a = b = c$.  
When $ABC$ is not equilateral, we find


*

*$\min_{X\in\mathbb{R}^2}(R_1R_2 + R_2R_3 + R_3R_1) = \min(ab,bc,ca) = bc$.

*If $a > b$, $X$ is unique and located at $A$.

*If $a = b > c$, $X$ can either be $A$ or $B$.


When $ABC$ is equilateral, it is clear $X$ can be $A,B,C$.
For $X \notin \{ A, B, C \}$, if the minimum is ever achieved, we need the inequality in $(*2)$ and the CS inequality become equality. This leads to two conditions
$$a' : b' : c' = x : y : z\quad\text{ and }\quad X = \frac{zA' + yB' + xC'}{x+y+z}$$
   By plugging in the values, it is easy to see first condition is
trivially satisfied while the second condition reduces to
$$\frac{\overrightarrow{XA}}{|XA|} + \frac{\overrightarrow{XB}}{|XB|} + \frac{\overrightarrow{XC}}{|XC|} = \vec{0}$$
which is satisfied at and only at the centroid $G$. 
As a summary, we have
$$\bbox[8pt,border:1px solid blue]{
\min_{X\in\mathbb{R}^2}\{ R_1R_2 + R_2R_3 + R_3R_1 \} = bc
\;\text{ achieved at }\;
\begin{cases}
A; & a > b\\
A, B; & a = b > c\\
A, B, C, G; & a = b = c
\end{cases}
}
$$
Notes


*

*$\color{blue}{[1]}$ M.S. Klamkin, Geometric inequalities via the polar moment of inertia, Mathematics Magazine,
48(1) (1975), 44–46.

*$\color{blue}{[2]}$ T. Hayashi, Two theorems on complex number, Tôhoku Math. J., 4 (1913-1914), 68–70.

*$\color{blue}{[3]}$ For the stronger inequality $(*3)$ by T. Hayashi, if $X$ is distinct from the vertices, equality holds when and only when $\triangle ABC$ is acute-angled and $X$ is the orthocenter.
A: Experiments (which I did using Cinderella) suggest that the value is minimal around one of your corner points. So to find the best value, compute your formula for all three corners and pick the minimal one.

Idea for a proof:


*

*Pick some coordinates without loss of generality, e.g. $A=(0,0),B=(1,0),C=(s,t),X=(x,y)$.

*Demonstrate that the partial derivatives become zero for $X\in\{A,B,C\}$.

*Show that the second derivatives actually indicate a local minimum.

*Show that there can be no other local minima which are even better.


The last one is probably the hardest.
A: I haven't solved this completely, but I found some facts:
First, each vertex is a local minimum, and any local minimum that's not a vertex must be inside the triangle (not on the edge, or outside).
Next, if $D$ is a point inside $\triangle{ABC}$, such that $D$ achieves a local minimum, then$$\frac{BD+CD}{\sin{\angle{BDC}}} = \frac{CD+AD}{\sin{\angle{CDA}}} = \frac{AD+BD}{\sin{\angle{ADB}}}$$
