Consider points $ E$ and $ D$, two of the points $ (A,B,C)$ lie in the same side of the line $ {ED}$wlog suppose they are $ A\&B$ then, one of the segment pairs $ (AD,BE)$ or $ (AE,BD)$ produces an intersection (Both segments intersect). wlog suppose $ AD\cdot BE=O$ because of triangle inequality: $ AO+OB>AB$ and $ EO+OD>ED$ $ \Rightarrow$ $ AD+BE>ED+AB$. On the other hand, $ AE+EC>AC$ and $ BD+DC>BC$ (Triangle inequality) adding up all inequalities: $ BD+CD+AE+EB+AD+CE>AB+AC+BC+ED$ as we wanted.

Image not found From the picture and from triangle inequality, we get: $ AE+EB>AB$ $ BD+DC>BC$ $ AF+FC>AC$ $ DF+FE>DE$ where F is the intersection of line segment $ AD$ and $ CE$ Adding up the inequalities we had, we get: $ AE+EB+BD+DC+AF+FC+DF+FE>AB+BC+CA+DE$ $ AE+BE+BD+CD+(AF+FD)+(CF+FE)>AB+BC+CA+DE$ $ \therefore AE+BE+BD+CD+AD+CE>AB+BC+CA+DE$ that is $ AB+BC+CA+DE<AD+AE+BD+BE+CD+CE$

If none of points A,B,C,D,E are on the same line than ABCDE is pentagono.We can easily prove that the BC+DE<BD+CE.We can take it from the cuadrangulo BCDE (sum of the diagonals is bigger than a sum of two oposite sides - prove easy).Than we can prove that AD+CD>AC and AE+BE>AB , it's because this sides are in one triangulos each. AB+BC+CA+DE<AD+AE+BD+BE+CD+CE Lets first put BC+DE<BD+CE : BD+CE+AB+AC<AD+AE+BD+BE+CD+CE AB+AC<AD+CD+AE+BE Lets put AD+CD>AC : AB+AC<AC+AE+BE AB<AE+BE Thats the right inequality (we had already proved that) so the starting inequality was also right.That means AB+BC+CA+DE<AD+AE+BD+BE+CD+CE So we are done. P.S I'm bad in English so if there are some mistakes in my speech dont mind and try to understand.

Since no three points are collinear, any three points make up a triangle. Draw a horizontal line segment $ DE$, with D to the left of E without loss of generality. Triangle $ ABC$ will be positioned with either 3 vertices above the line or 2 vertices above and 1 below, and the reflections of both of these cases. In any case 2 vertices are above or 2 are below, so without loss of generality, let A and B always be on the same side of the triangle. Note all possible positions of $ \triangle ABC$ have been considered. Without loss of generality, let A and B be positioned above the triangle, and C anywhere. There are two possibilities: B is inside triangle ADE or A is inside triangle BDE (1 possibility), or B is outside triangle ADE and vise versa (again, note there are no other possibilities). In the first case, where without loss of generality B is inside $ \triangle ADE$, $ AD + DC > AB + BC$ (or $ AE + EC > AB + BC$, depending on whether C is to the right or left of line AB). Then $ AE + CE > AC$ (or $ AD + CD > AC$), and $ BD + BE > DE$, so putting this all together yields the desired result. For the second possibility, if B is not inside triangle ADE and A is not in triangle BDE, then two of the segments on the greater than side of the inequality must intersect each other. Without loss of generality, let A be to the left of B (or more exactly, there exists a line normal to DE to which A is left of and B right of). Note if neither is to the left of the other then we have the previous possibility occuring. Then AE and BD (and it must be these because of the second sentence of the problem) cross each other at a point we'll creatively call F. So $ AE = AF + FE$ and $ BD = BF + FD$. Note no restrictions have been placed on C (besides existence); it could be anywhere and the reasoning of this explanation is in no way affected. Then the inequality is a conglomeration of the following triangle inequalities: $ AF + BF > AB$, $ FD + FE > DE$, $ CD + AD > AC$, and $ CE + BE > BC$.