For set of segments, check if they can create n-side polygon If we have given set of $n$ segments (or lengths of segments), how can we easily check if this set is forming polygon with $n$ sides.
Example
Let our set be $A$, $A = \{3, 6, 6\}, n = 3$
The answer here should be true, because we can arrange those segments to make triangle.
But the problem for me is when there are more segments.
So I tried to come with some statment, I started from triangle, it says: To form a triangle there shouldn't be side that is bigger than the sum of the other two sides, and I came to statement for more than 3 sides.
Is this correct: For any n-sided polygon, there shouldn't be side that is bigger than the sum of any two other sides.
Thanks in advance.
 A: You are on the right track, but the correct statement is that $n$ lengths admit a (convex) polygon if and only if no one length is greater than or equal to the sum of the other lengths.
To sketch a proof of necessity, consider any side $E$ of a polygon.  The other sides form a path from one endpoint to the other, and since the shortest distance between the endpoints is given uniquely by the straight line segment (edge length), the sum of remaining lengths exceeds that of edge $E$. 
We can also say a little about a proof of sufficiency, e.g. the construction of a polygon given side lengths, no one of which equals or exceeds the sum of the remaining lengths.  It is possible to proceed by induction from the base case $n=3$ of triangles.  The previous Question on maximizing the area of such constructed polygons shows a more elegant result.   
This is a straightforward generalization of the case for triangles, and again it suffices to check that the longest segment's length is less than the sum of the remaining lengths.
