Triangles and inequalities

I'll use $P$ instead of $O$ through the course of this answer; $O$ is a little confusing because it usually denotes a centre.

Let $D$, $E$ and $F$ be midpoints of $BC$, $CA$, and $AB$ respectively; let $G$ be the intersection of $AD$, $BE$, and $CF$, i.e. the triangle's centroid. It is well known that $AG$ = $2GD$ (and similar for the other medians). By Apollonius's theorem, along with the fact that $AD = \frac 3 2 AG$, we note that $$ \frac 9 2 AG^2 + \frac 1 2BC^2 = AB^2 + AC^2; $$ summing the symmetric equations for the other medians we have $$ \frac 9 2 (AG^2 + BG^2 + CG^2) = \frac 3 2 (AB^2 + BC^2 + CA^2). $$ In other words, $G$ makes the inequality tight. Thus we are clearly motivated to try and do something with $G$.

Consider the following diagram.

By Stewart's theorem (a slight generalisation of Apollonius's theorem, and only a few steps removed from direct Pythagoras bashing) we obtain the result that $$ \frac 2 3 PD^2 + \frac 1 3 PA^2 = PG^2 + \frac 2 9 AD^2 = PG^2 + \frac 1 2 AG^2. $$ To get rid of the $PD^2$, we use Apollonius on $\bigtriangleup \!\! PCB$: $$ 2PD^2 + \frac 1 2 BC^2 = PB^2 + PC^2. $$ Multiplying the former expression by $3$ and subtracting the latter we obtain $$ PA^2 - \frac 1 2 BC^2 = 3PG^2 + \frac 3 2 AG^2 - PB^2 - PC^2. $$ Summing cyclically for the corresponding expressions oriented about medians $BE$ and $CF$, and substituting in our previous expression for the sum of square distances at the centroid $G$, we obtain $$ 3(PA^2 + PB^2 + PC^2) = 9PG^2 + AB^2 + BC^2 + CA^2, $$ so we are done by non-negativity of squares.