Here is an interesting question that I have been thinking about for awhile now but do not know the answer to.
Suppose you have a convex polygon with $n$ sides. What would be an example of such a polygon s.t. if you randomly picked three sides of the polygon, they could not form a triangle? I am talking about an $n$ sides polygon not some polygon with sides like $5$ or $10$. Just $n$ sides.
Most polygons that people draw will have all the sides about the same length, and any set of three sides can form a triangle. I think it is more interesting to ask what polygons have some set of three sides that cannot form a triangle.
One kind that has a set that cannot form a triangle is a single long side, say $10$ units long, with lots of $1$ unit sides to connect the ends. If your group of three has the long side in it, you can't form a triangle. You can also have some number of long sides (of differing lengths) connected by lots of short sides. Then if you take two longs and a short you also can't get a triangle.
Added, you can make a quadrilateral, even a trapezoid, with sides $3,1,2,1$ where no three sides will form a triangle. Then a pentagon $5,3,1,2,1$ and generally an $n-$gon with the Fibonacci series.
If the sides are $s_i$ with $s_i\le s_{i+1}$, we need $s_{j-2}+s_{j-1} \le s_j \lt \sum_{k=1}^{j-1}s_k$ You can pick $s_1,s_2$ as you wish (and might as well scale to make $s_1=1$), then respect the left inequality for $s_3$, then get polygons of arbitrary size by successively picking the sides. Once you get started, there will always be room for the next side.
Added again: For a geometric series, you need $1+r \le r^2$, which says $r \ge \phi$ In fact you can use $\phi^n$, which is just about the Fibonacci series. Then you need $r^n\lt \frac {r^n-1}{r-1}$. This seems to be most trying at $n=3$, restricting us to $r \lt \frac 13 (1+(19-3 \sqrt{33})^{1/3}+(19+3 \sqrt{33})^{1/3}) \approx 1.839286$ as found by Alpha. I checked it in Excel and it works at $1.83928$