Conditions for formation of an $n$ sided polygon 
Given $n$ sides of any arbitrary length, give the necessary and sufficient conditions such that they form a $n$ sided polygon. 

For a triangle, if the sides are given by $a\le b\le c$ then the necessary and sufficient condition will be $a+b > c$, that is sum of two smaller sides should be greater than the third. This result can be derived by calculating the area of this triangle by Heron's formula. 
For a general $n$ sided polygon, let the sides in incresing order be $a_1 \le a_2 \le \cdots \le a_n$, then a necessary condition will be $a_n < a_1+\cdots+a_{n-1}$. Consider the distance between the endpoints of side $a_n$. There are two different paths connecting them, one path has the length $a_n$ and other one has $a_1+\cdots+a_{n-1}$, since the shortest distance between two points is a straight line between them, we get our required result. But will this also be sufficient? Maybe $n=4$ case can help us understand the general case.  
$\hskip2in$ 
 A: We are going to prove that provided the lengths of $n$ segments ($n\ge3$) satisfy the condition:
$$
0<a_1\le a_2\le\dots\le a_n;\quad
a_n<\sum_{i=1}^{n-1} a_i
$$ 
it is possible to construct out of these segments a $n$-sided polygon.
I will assume that for triangles ($n=3$) the statement is established (let me know if you have troubles with this and I will provide a proof). 
Assume now that the statement is valid for $n-1$ segments. We will demonstrate that this implies that it is valid for $n$ segments as well (observe that $n$ is at least $4$).
Consider the following cases:


*

*$a_n-a_1>a_{n-1}: \max(a_n-a_1,a_{n-1},\dots,a_2)=a_n-a_1;\quad a_n-a_1< \sum_{i=2}^{n-1}a_i$;

*$a_n-a_1\le a_{n-1}: \max(a_n-a_1,a_{n-1},\dots,a_2)=a_{n-1};\quad a_{n-1}< a_n-a_1+\sum_{i=2}^{n-2}a_i$.
Thus if we decrement the length of the longest segment by that of the shortest one and remove the shortest segment from the set, the longest of the remaining segments will be less than the sum of the rest $n-2$ segments. 
Let $AB=a_{n}$ be the longest segment. Construct a point $F$ on the segment $AB$ such that $AF=a_1$ (see the figure below). By the induction hypothesis from the remaining $\color{red}{n-1}$ segments we can construct a polygon $BC\dots EF$. Now reflect the segments $AF$ and $FE$ about the line $AE$ to obtain the polygon $ABC\dots EF'$ with $\color{red}n$ sides.

PS. One can note that the above proof fails in the case 2. if (and only if) $n=4$ and the equalities $$a_1=a_2,\quad a_3=a_4$$ hold. In this case the explicit construction of the quadrilateral (parallelogram or kite) helps.
