, 2009

Let $G$ denote the centroid of $ABC$. We have
$$\sum \frac{PA}{a} = \sum \frac{PAGA}{aGA} \ge \frac{\sum PAGA}{\max\{aGA, bGB, cGC\}}.$$
We observe that $\vec{PA} \cdot \vec{GA} \le PAGA$. Thus
$$\begin{align*} \sum PAGA & \ge \sum \vec{PA} \cdot \vec{GA} \\ &= \sum (\vec{PG} + \vec{GA}) \cdot \vec{GA} \\ &= \vec{PG} \cdot (\vec{GA} + \vec{GB} + \vec{GC}) + \sum GA^2 \\ &= GA^2 + GB^2 + GC^2. \end{align*}$$

$$\begin{align*} \sqrt{3}am_a &= \frac{\sqrt{3}}{2}a(2m_a) \\ &= \frac{\sqrt{3}}{2}a\sqrt{2(b^2+c^2)-a^2} \\ &\le \frac{1}{4}(3a^2+2b^2+2c^2-a^2) \\ &= \frac{1}{2}(a^2+b^2+c^2). \end{align*}$$
This implies that $3\sqrt{3}aGA \le a^2 + b^2 + c^2$. Hence we obtain
$$\sqrt{3}aGA \le \frac{1}{3}(a^2 + b^2 + c^2) = \frac{4}{9} \sum m_a^2 = \sum GA^2.$$
Similarly, we obtain
$$\sqrt{3}bGB \le \sum GA^2, \quad \sqrt{3}cGC \le \sum GA^2.$$

It follows that
$$\max\{aGA, bGB, cGC\} \leq \sum GA^2.$$
Therefore
$$\sum \frac{PA}{a} \geq \frac{\sum GA^2}{\sum GA^2 / \sqrt{3}} = \sqrt{3}.$$