Triangle $ABC$ has a circumcircle $(O;R)$. $M$ is a point in triangle $ABC$. Prove that $$MA^{2}+MB^{2}+MC^{2}\geq \frac{4S_{ABC}}{\sqrt{3}}\left(1+\frac{OM^{2}}{3R^{2}}\right)$$
When I prove I think lots of ways to start. I see $\dfrac{OM^{2}}{R^{2}}$, I think we can use Euler theorem about the pedal triangle. With $MD,ME,MF$ is perpendicular to $BC,CA,AB$ we have $$S_{DEF} = \frac{1}{4}\left(1-\frac{OM^{2}}{R^{2}}\right).S_{ABC}$$
Besides that, I also think we can use the lemma$$ \frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c}\geq \sqrt{3}$$
Can anyone help me, please?