<s>[INST] 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}|$$ is as small as possible.
 [/INST] Let $a = |BC|$, $b = |CA|$, $c = |AB|$, $R_1 = |XA|$, $R_2 = |XB|$, $R_3 = |XC|$.  
WOLOG, we will assume $a \ge b \ge c$. We will show that
$$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.
</s>