Proving that the distance between a point and the median intersection point is $\le$ than 1/3 the sum of distances between the point and vertexes

Question

Given a triangle $ABC$. The point $M$ is the median intersection point. $O$ is an arbitrary point. How can I prove that $$OM \le \frac{OA + OB + OC}{3}?$$

I think that first we need to prove that $\vec{OM} = \vec {OA} + \vec{OB} + \vec {OC}$.

Answer 1

Sorry, I really am. The problem is solved this way: $$\vec{OM}=\vec{OA}+\vec{AM}$$ $$\vec{OM}=\vec{OB}+\vec{BM}$$ $$\vec{OM}=\vec{OC}+\vec{CM}$$

So $\;3\cdot \vec{OM}=\vec{OA}+\vec{OB}+\vec{OC}+(\vec{AM}+\vec{BM}+\vec{CM})$.

Now $\vec{AM}+\vec{BM}+\vec{CM}=0\,.\;$ It's all trivial. Sorry for misleading you. (I deleted a bad idea that overcame me). So we get this $\;3\cdot \vec{OM}=\vec{OA}+\vec{OB}+\vec{OC}\,,\;$and $\;3\cdot OM\leqslant OA+OB+OC\,.$

PS for example, it's obvious that $\vec{AM}=1/3(\vec{AC}+\vec{AB})\,.\;$ And we get similar expressions for vectors $\,\vec{BM}\,$ and $\,\vec{CM}\,.\,$ They then will all cancel out when we add them up $\vec{AC}+\vec{CA}=0$ etc. In other words we get:

$\vec{BM}=1/3(\vec{BA}+\vec{BC})\,,\quad$ $\vec{CM}=1/3(\vec{CA}+\vec{CB})\,.\quad$

Hence, $\;\vec{AM}+\vec{BM}+\vec{CM}=\frac{1}{3}(\vec{AC}+\vec{AB}+\vec{BA}+\vec{BC}+\vec{CA}+\vec{CB})=0\,.$

This formula $\,\vec{AM}=1/3(\vec{AC}+\vec{AB})\,$ is derived using the fact that medians are intersected at the point called centroid which divides each of the medians in the ratio $2:1$. You can also extend the triangle and get a parallelogram to make things very trivial.

Answer 2

$$\sum_{cyc}\vec{AM}=\sum_{cyc}\left(\frac{2}{3}\cdot\frac{1}{2}\vec{AB}\right)=\frac{1}{3}\sum_{cyc}\vec{AB}=\vec{0}.$$ Thus, $$\sum_{cyc}\vec{OA}=\sum_{cyc}\left(\vec{OM}-\vec{AM}\right)=3\vec{OM},$$ which says $$OM=\left|\frac{1}{3}\sum_{cyc}\vec{OA}\right|\leq\frac{1}{3}\sum_{cyc}\left|\vec{OA}\right|=\frac{OA+OB+OC}{3}$$ and we are done!