Sum of Triangle Vertices Distances from an Internal Point

Let $AB=c$ , $AC=b$ , $BC=a$ , $OC=f$ , $OA=g$ , $OB=h$ and extend $OC$ to a point D as shown in the figure above. so $P=a+b+c$ and $S=f+g+h$ .

First we will prove that the minimal value of $S$ is $P/2$ :

Using the theorem that the sum of any two sides in a triangle is larger than the third side we can say:

In $\triangle AOB$ $\longrightarrow$ $g+h > c$

In $\triangle AOC$ $\longrightarrow$ $f+h > a$

In $\triangle BOC$ $\longrightarrow$ $f+g > b$

Summing the three inequalities will yield:

$2f+2g+2h>a+b+c$
$2(f+g+h)>a+b+c$
$2S>P$

$\boxed {S>P/2}$

Now we will prove that the maximal value of $S$ is $P$ :

This one is a bit trickier, we will use the extension of $CO$ to $CD$ and remember that $c=AD+DB$

Using again the same theorem in $\triangle OAD$ we get:

$g < OD+AD$
$g+OC < OD+AD+OC$
$(1)$ $g+f < CD+AD$

Using again the same theorem in $\triangle CBD$ we get:

$CD < a+DB$

$CD +AD < a+DB+AD$

$CD +AD < a+DB+AD$

$(2)$ $CD +AD < a+c$

Adding inequalities $(1)$ and $(2)$ we will get:

$(3)$ $g+f < a+c$

note that this is a general proof so if we extend $OA$ and $OB$ we can repeat the same process and yield:

$(4)$ $f+h < b+c$

$(5)$ $g+h < a+b$

Adding inequalities $(3)$ , $(4)$ and $(5)$ will yield us:

$2f+2g+2h < 2a+2b+2c$

$f+g+h < a+b+c$

$\boxed{S <P}$

Therefore overall:

$\boxed{P/2<S <P}$

Would be happy to see your solutions to the problem :)