How prove $\sum\limits_{cyc}\sqrt{PA+PB}\ge 2\sqrt{\sum\limits_{cyc}h_{a}}$

Question

Question:

Consider a triangle $\Delta ABC$ with altitudes $h_{a}$, $h_{b}$ and $h_{c}$, where $AB=c$, $BC=a$ and $AC=b$. Show that for any $P$

$$\sqrt{PA+PB}+\sqrt{PB+PC}+\sqrt{PA+PC}\ge 2\sqrt{h_{a}+h_{b}+h_{c}}$$

My try: the inequality is equivalent to

$$(PA+PB+PC)+\sum_{cyc}\sqrt{(PA+PB)(PB+PC)}\ge 2(h_{a}+h_{b}+h_{c})$$

Answer 1

I noticed that the LHS is dependent on $P$ but the RHS isn't. So why not try the following approach: Find out the $P$ for which the LHS is minimum , and prove the inequality. Then we are done!

For any given triangle $ABC$, the minimum value of LHS will occur if the $P$ is the Circumcentre. i.e. $PA=PB=PC=R\ (circumradius)$. The inequality reduces to
$$ 9R\ge 2(h_a+h_b+h_c)$$ Now we know $R,h_a,h_b,h_c$ in terms of the sides and angles. Can you proceed from here?