Let $ABC$ be a triangle on the Euclidean plane. At which point $P$ on the plane does the ratio sum $\frac{PA}{BC}+\frac{PB}{CA}+\frac{PC}{AB}$ attain its minimum value? Prove also that, for any point $P$ on the plane, $$\frac{PA}{BC}+\frac{PB}{CA}+\frac{PC}{AB} \geq \sqrt{3}\,.$$ When does the equality hold?
Source: Problem 965 from http://www.imomath.com/othercomp/Journ/ineq.pdf (Crux Problem by George Tsintifas, Thessaloniki, Greece).
EDIT: The second part of the problem has been solved in the link given in a comment below. Please attempt only the first part (i.e., finding the minimizing point).
As I said in my comment above, one can define: $\alpha=1/BC$, $\beta=1/AC$, $\gamma=1/AB$ so that we are required to find the point $P$ such that $\alpha PA + \beta PB + \gamma PC$ attains its minimum value. Following the method outlined here let's denote by $\vec{i}$, $\vec{j}$ and $\vec{k}$ the unit vectors along $\vec{PA}$, $\vec{PB}$ and $\vec{PC}$ respectively; if $X$ is another point, one finds: $$ \tag{1} \alpha PA + \beta PB + \gamma PC\le \alpha XA + \beta XB + \gamma XC +\vec{PX}\cdot(\alpha\vec{i}+\beta\vec{j}+\gamma\vec{k}). $$
It follows that if $\alpha\vec{i}+\beta\vec{j}+\gamma\vec{k}=0$ then $P$ is the minimum point. That condition is equivalent to $\vec{k}=-(\alpha\vec{i}+\beta\vec{j})/\gamma$ and this can be true only if $(\alpha\vec{i}+\beta\vec{j})^2=\gamma^2$, that is if \begin{equation} \tag{2} \cos(\angle APB) = \vec{i}\cdot\vec{j}={\gamma^2-\alpha^2-\beta^2\over 2\alpha\beta}, \end{equation} which can be satisfied if the absolute value of the right hand side does not exceed $1$. In that case we also have two other analogous equations for $\cos(\angle BPC)$ and $\cos(\angle CPA)$, so we can find those three angles and locate $P$.
If, on the other hand, the absolute value of the right hand side of $(2)$ is greater than $1$, suppose $\gamma\ge\alpha$ and $\gamma\ge\beta$ (i.e. $AB\le BC$ and $AB\le AC$). Then we have from $(2)$: $\gamma>\alpha+\beta$ and, redefining $\vec{k}=-(\alpha\vec{i}+\beta\vec{j})/\gamma$ and taking $P=C$, equation $(1)$ still holds (because $|\vec k|<1$). In that case the requested minimum point is then $C$, that is the point opposite to the shortest side.