When $\triangle ABC$ is right-angled, the inequality is trivial, thus we assume that $\triangle ABC$ is acute. Let $D,E,F$ denote the midpoints of $BC,CA,AB.$ Since $O$ becomes orthocenter of $\triangle DEF$ (also acute) then $O$ is inside of it $\Longrightarrow$ $O$ is inside the quadrilateral $BFEC,$ thus $BF+FE+EC>BO+OC$ $\Longrightarrow$ $a+b+c>4R$ $\Longrightarrow$ perimeter of $\triangle ABC$ is greater than two times its circumcircle diameter.