Prove for Arbitrary Point $O$ inside $\triangle ABC$ we have $OA + OB + OC < AB + AC + BC$

Question

Prove for Arbitrary Point $O$ inside $\triangle ABC$ we have $OA + OB + OC < AB + AC + BC$.

I know $\frac{1}{2} (AB+AC+BC)<OA+OB+OC$ but don't know how to prove $OA + OB + OC < AB + AC + BC$.

Answer 1

Let $P\in \overrightarrow{AB}$ such that $O\in\overrightarrow{PC}$.

From triangle $PBO$, $OB < BP + PO$. Adding $OC$ both sides, $OB + OC < BP + PO + OC = BP + PC$. From triangle $APC$, $PC < AP + AC$, adding BP both sides, $BP + PC < BP + AP + AC = AB + AC$. Hence, $OB + OC < BP + PC < AB + AC$.

Similarly, $OA + OB < CA + CB$ and $OA + OC < BA + BC$. Adding all the inequalities, $2(OA + OB + OC) < 2(AB + BC + CA) \Rightarrow OA + OB + OC < AB + BC + CA$.

Answer 2

If $k= OA+OB+OC,\ l=AB+BC+AC$ then define a function $ p( \triangle XYZ)=XY+YZ+XZ$ So $$ p(\triangle OAB)+p(\triangle OBC)+p(\triangle OAC) \leq 3p(\triangle ABC) $$

That is $ 2k+l\leq 3l$ so that $k\leq l$

Answer 3

So, there's a basic lemma that if we have a triangle $XYZ$ and a point $T$ inside it, then the perimeter of triangle $XYT$ is less than the perimeter of $XYZ$. More generally, a convex figure containing another convex figure has larger perimeter.

If we assume this lemma, in our case it tells us that $AB+AC>OB+OC, BA+BC>OA+OC, CB+CA > OA+OB$ and when we add up and cancel the 2s, we get what we want.

For a more motivated way to see why the lemma is true, imagine the set of points for which $XT+YT$ is constant: it's an ellipse with foci $X$ and $Y$. Then it must intersect the sides $XZ,YZ$ at interior points. If it intersects say $XZ$ at $W$, we know $XT+YT = XW+WY< XW+WZ+ZY =XZ+YZ$ by the triangle inequality.

It's possible to prove the lemma using some calculations too, but that's not as illuminating.

EDIT: I was being dumb, there's a far easier way to prove the lemma: just look at Weberson Arcanjo (https://math.stackexchange.com/users/308563/weberson-arcanjo)'s answer