Is there a simple way to find a point $P$ in a $2\mbox{D}$ plane that minimizes $$\alpha(PA^2+PB^2+PC^2)+\beta(PA+PB+PC),$$ where $\alpha,\beta>0$? If $\alpha=0$, the answer is the Fermat point of $\triangle ABC$, while if $\beta=0$, the answer is the center of gravity of $\triangle ABC$. For general $\alpha,\beta$, how do I express the point $P$ (or vector $\overrightarrow{OP}$ from the origin $O$) in terms of $\,\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\overrightarrow{OC}$?
I tried putting everything in coordinates and find that $\,PA+PB+PC\,$ gives sum of square roots, which is not easy to minimize. The result also looks messy. So I hope to find a neat answer with the help of planar vectors.
One possible approach would be to use Cauchy-Schwartz $$\Big(\alpha\sum_{A,B,C}PA^2+\beta\sum_{A,B,C}PA\Big)\Big(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\Big)\geq\Big(\sqrt{\sum_{A,B,C}PA^2}+\sqrt{\sum_{A,B,C}PA}\Big)^2$$ then try to see if the point at which the latter is minimized could be expressed nicely. Obviously, there exists such a global minimum and thus a choice of $P$ gives the minimal value, but I sense that you are more interested in knowing if such a choice of $P$ is expressed nicely.