Geometrical proof for $PA+PB+PC\le3R$, where $P$ is the orthocenter and $R$ is the circumradius

Question

$ABC$ is an acute angled triangle, where $P$ is the orthocenter, and $R$ is the circumradius. I want to show that $PA+PB+PC\le 3R$ geometrically, that is without using trigonometry. I have a trig solution, but I want to know whether we can do it by pure geometry.

Note: In the image, the direction of the inequalities should be the opposite.

Answer 1

Hint: the identity $a^2 + b^2 + c^2 + PA^2 + PB^2 + PC^2 = 12R^2$ is useful.

Answer 2

Yes, there is a way. Due to the Euler theorem, $O,G,H$ are collinear and $HO=3\,GO$.

This implies that if we take $O$ as the origin, the vector equation: $$ H = A+B+C $$ holds. By applying the triangular inequality: $$ OH = |H| \leq |A|+|B|+|C| = OA+OB+OC = 3R.$$