How to prove $\sum_{i PA_1+PA_2+....+PA_n$ 
How to prove two inequalities as follows:

Let $A_1A_2....A_n$ be convex polygon, $P$ be a point inside of $A_1A_2....A_n$.
$$A_1A_2+A_2A_3+....+A_{n-1}A_n > 2\frac{PA_1+PA_2+....+PA_n}{n-1}$$
$$\sum_{i<j} A_iA_j > PA_1+PA_2+....+PA_n$$
See: Triangle inequality
 A: Prove this yourself :

If a convex polygon is completely contained inside another polygon, then the perimeter of the first is smaller than the perimeter of the second.

Use induction, and triangle inequality.
Once this is done, apply this to the triangle $PA_iA_{i+1}$ (where $i+1$ will return to $1$ if $i$ is the last vertex) and the whole polygon. Each one will give you $PA_i + PA_{i+1}  + A_iA_{i+1} < S$, where $S$ is the perimeter of the convex polygon.
Adding these up, we get $2\sum PA_{i} < nS - S < (n-1)S$, which was to be proved.
For the second, fix a vertex $A_j$. Then, note that for $i \notin \{j-1,j+1\}$, we have that $A_{i}A_{i+1} < A_iA_j + A_{i+1}A_j$. Summing over all such $i$, and adding $A_{j-1}A_j$ and $A_{j}A_{j+1}$ to both sides, we get that $\sum_{i \neq j} A_jA_i > S$. Now, sum over all $j$, to get $2\sum_{i \neq j} A_jA_i > nS$. However, we know that $nS > (n-1)S > 2\sum PA_i$. Cancelling out both sides and dividing by $2$ gives that $\sum_{i < j} A_iA_j > \sum PA_i$, which was to be proved.
