This is a slight variation on the usual broken stick problem.
A stick is broken randomly into two pieces. The larger piece is then broken in two. What is the probability the pieces can form a triangle?
I take the stick as the interval $(0,1)$ and the first break as a uniform random variable $X$. Then let $Y$ be a uniform random variable on $(0,X)$. It suffices to look at $X≥1/2$ and multiply the answer by $2$. Now we want to compute the probability that $Y≤1/2$ and $X−Y≤1/2$.
My problem is that I get a probability involving $X$ so I am clearly doing something wrong, specifically I end up with $\frac{1}{4 X^2}$ ...
The diagram below shows the relevant regions in the $X$-$Y$ plane. We uniformly choose $X\gt\frac12$ and then choose $Y\lt X$, yielding the shaded region. The region with the darker shade is the part where both $Y\lt\frac12$ and $X-Y\lt\frac12$. Given $X=x$, the proportion of the darker shade is $\frac{1-x}x=\frac1x-1$ (as Henry noted in a comment). Averaging this over $x$ yields the probability
$$ 2\int_\frac12^1\left(\frac1x-1\right)\mathrm dx=2\left[\log x-x\right]_\frac12^1=\log4-1\approx39\%\;. $$