prove that sum of lengths of sides of pentagon is less than sum of lengths of diagonals of pentagon 
Let $ABCDE$ be pentagon. Prove that sum of lengths of sides of pentagon is less than sum of lengths of diagonals of pentagon

APPROACH 1
I tried using triangle inequality but it does not lead to a proof.One thing i noticed that the statement is not true for quadrilaterals.I proved some extreme cases like when one vertex is collinear with two other vertices.
APPROACH 2
If we consider sides of pentagon as vectors,then diagonals are just vector sum of sides.I thought this might help. I ended with the following inequality to prove
$\vert{\vec a}\vert+\vert{\vec b}\vert+\vert{\vec c}\vert+\vert{\vec d}\vert+\vert{\vec e}\vert < \vert{\vec a + \vec b}\vert+\vert{\vec c + \vec b}\vert+\vert{\vec c + \vec d}\vert+\vert{\vec d + \vec e}\vert+\vert{\vec e + \vec a}\vert$
Again i stuck here.I want to ask whether my approaches can actually lead to a proof or i may ness some other approach?
 A: Let F be the intersection of AC and BD.  Then $ BF+CF\gt BC$.  etc.
It works for a regular pentagon; I'm not certain that, for example, BF+GD is always less than BD.
EDIT:
Take the nonconvex pentagon  $$A(0;0),B(1,10),C(2,1),D(3,10),E(4,0)$$
The sides have length over$40$, but the diagonals not much more than $20$.
A: We'll show this for convex pentagons.
$\bf{Added:}$ Oh, the solution is embarassingly simple

The sum of diagonals equals the perimeter of the star decagon $+$ the perimeter of the inside small pentagon. But it's easy to see that the already the perimeter of the star decagon is larger than the perimeter of the pentagon ( use the triangle inequality, two sides of the decagon $>$ a side of the large pentagon).
$\bf{Alternate\ \  solution:}$  ( an overkill)
Let us use the Crofton's formula
$$l(\gamma) = \int_{l} n(\gamma, l)\, d\!\mu(l)$$
where $\gamma$ is a rectifiable curve, $l$ is a line in the plane, $\mu$ is a certain measure on the space of lines, and $n(\gamma, l)$ is the number of times the line $l$ intersects the curve $\gamma$.
Now let $\gamma_1$ be the closed curve formed by the pentagon, and $\gamma_2$ be the closed curve formed by the diagonals of the pentagon ($\gamma_2$ is a pentagram).
Now let us note that whenever $l$ is a line such that $n(\gamma_1, l)> 0$ (being $1$ or $2$), we have
$$n(\gamma_2, l) \ge n(\gamma_1, l)$$
Now use the Crofton formula to get the inequality.
Note: Crofton's formula for a convex closed curve can be interpreted as follows: the length of a closed convex curve  is proportional to the average  length of the projection of the curve along different directions.
$\bf{Added:}$
Note: The inequality may not be true for a concave pentagon. As example, consider a concave pentagon with vertices approximately $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(1,0)$, $E(2,0)$.
