I'm not any good at math, nor do I understand much about probability, but I've always approached this problem from graphically.
Lets say the length of the stick is 1 and assume that the cuts on the sticks are distributed uniformly. Lets call the distance from the left end of the stick to the first cut $x$, and the distance from the stick to the second cut $y$, and lets assume that the stick is of length 1:

We know that the sum of any two sides of a triangle must be greater than the third side. Thus we have:
$$x + (y-x) > 1-y$$
$$(y-x)+(1-y)>x$$
$$x+(1-y)>y-x$$
And simplifying:
$$y>.5$$
$$.5>x$$
$$x+.5>y$$
Let us also add the condition that the $y$ is the first plus second piece starting from the left, and x is always the first piece, so:
$$y>x$$
We also know that the most the $x$ piece can be in length is 1, and the least is 0, same with y, so we add conditions:
$$1>x>0$$
$$1>y>0$$
If you graph out the last 3 conditions I gave then that graphically will give you the whole area where it is possible for you to get when you do the two cuts, and then the first 3 conditions I gave will give you the piece of that area that allows you to make a triangle. I don't know how to draw a graph, but you should get a smaller triangle that is 1/4 the bigger triangle, meaning .25 chance of getting a triangle with 2 uniformly distributed cuts.