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