Triangles and inequalities Let ABC be a triangle and let O be any point in space. How can I show that  $AB ^ 2 + BC ^ 2 + CA ^ 2 \leq 3 (OA ^ 2 + OB ^ 2 + OC ^ 2)$?
I know this prove by inner products, but is possible to show this by euclidean geometry?

If we find such point $D$ that $$3(OA^2+OB^2+OC^2)=3(DA^2+DB^2+DC^2+3OD^2)\tag{1}$$ then we will only have to show that $$AB^2+BC^2+CA^2\le 3(DA^2+DB^2+DC^2)\tag{2}$$
As I'm too lazy to prove this geometrically about sufficent choice of $D$ as median intersection point (i.e. centroid) of $\triangle ABD$, let's put the entire thing into Cartesian coordinates, letting $A(a,b),\,$ $B(c,d),\,$ $C(u,v),\,$ $O(x,y),\,$ $D(0,0)$, then $$OA^2+OB^2+OC^2-(DA^2+DB^2+DC^2+3OD^2)=\\
-2 (a x + b y + c x + d y + u x + v y)$$
so we select $D$ and move the origin into it such a way, that 
$$a+c+u=0,\,b+d+v=0$$
And now we have to prove $(2)$ with this choice of $D$, as we have $(1)$, but in this case the equality holds $$AB^2+BC^2+CA^2=3(DA^2+DB^2+DC^2)\tag{3}$$
so we're done.
Update: also the equalities $(1)$ and $(3)$ are given on the wiki (Centroid) with $O$ being $P$ and $D$ being $G$ there.
