Will $n$ segments form a polygon IFF the sum of $n-1$ sides is greater than the remaining side?

Question

I know the IFF condition holds for a triangle, but does the IFF condition hold true for any $n$-sided polygon?

The IFF condition is clear in the forward direction in that it's intuitive why any $n$-sided polygon would require that the sum of any $n-1$ sides be greater than the remaining side. But if the IFF condition is true, it's not intuitive to me why the sum of any $n-1$ sides being greater than the remaining side is a sufficient condition to ensure an $n$-sided polygon.

Answer 1

For an intuitive explanation, imagine that the side are thing hollow metal rods, with a chain running through all of them and joining them. (The chain forms a closed loop.)

Hold the longest rod parallel to the ground. The other rods hang down an close the polygon.

Answer 2

If there are 4 or more sides, you can always reduce the number of sides by joining together two adjacent ones into one long edge (essentially setting the vertex angle to 180). If that would result in a side longer than all the other sides put together, simply change your mind and choose an adjacent pair of those other sides instead (which when joined together clearly cannot form a side that is longer than the sum of the others). Repeat till you have only 3 sides left.

So if you can see why it is sufficient for a triangle, it is clear that it works for the original sequence of side lengths, as they form a triangle with some extra vertices on the sides of the triangle.

Answer 3

This can be rigorously proved by induction (which may or may not be intuitive to you).

I assume the result is clear for $n=3$, i.e. the triangle (let me know if it's not). For $n \geq 4$, let $l_1 \leq \cdots \leq l_{n-1} \leq l_n$ denote the sequence of lengths. Now let $$ \epsilon = \frac{1}{2}\min_j\left\{\sum_{i=1}^n l_i - 2l_j, l_1 \right\}, $$ and note $\epsilon > 0$ as it is a minimum of positive numbers (where positivity comes from the assumption on the $l_i$'s). Consider the sequence $a_1 = l_1 + l_2 - \epsilon, a_2=l_3, \cdots, a_{n-1}=l_n$; note $a_1 = l_1 + l_2 - \epsilon > l_2 > 0$. Thus we have a collection of $n-1$ lengths. Moreover, every sum of $n - 2$ of these is greater than the remaining length. Indeed, if $a_1$ is in the sum, we have $$ \sum_{\substack{1 \leq i \leq n - 1 \\ i \neq j}}a_i = \sum_{i = 1}^{n-1} a_i - a_i = \sum_{i = 1}^n l_i - l_{j+1} - \epsilon > l_{j+1} = a_{j} $$ because $\epsilon < \sum_{i=1}^n l_i - 2l_{j+1}$. For the remaining sum, first note $l_1 + l_2 \leq \sum_{i=3}^{n} l_i$; indeed, because $n \geq 4$ we have $l_1 \leq l_3$ and $l_2 \leq l_4$. Now we may compute $$ \sum_{i=2}^{n-1}a_i = \sum_{i=3}^n l_i \geq l_1 + l_2 > l_1 + l_2 - \epsilon = a_1 $$ Hence by induction we see that $a_1, a_2, \cdots, a_{n-1}$ are the side lengths of some polygon. Finally, note $l_1, l_2, l_1 + l_2 - \epsilon$ are the side lengths of a triangle (verify at your leisure). Thus we may attach the triangle with these side lengths (identifying the sides of length $l_1 + l_2 - \epsilon$) to our newly-constructed polygon to obtain a polygon with side lengths $l_1, l_2, \cdots, l_n$ and complete the proof!