What is the probability to form a triangle with the three pieces of the stick? 
On a stick $1$ meter long is casually marked a point $X \sim U[0,1]$. Let $X=x$, is also marked a second point $Y\sim U[x,1]$.


1) Find the density of $(X,Y)$ showing the domain.

$$\rightarrow \quad f_{XY}(x,y)=\frac{1}{1-x}\mathbb{I}_{[0,1]}(x)\mathbb{I}_{[x<y<1]}(y)$$

2) Say if $X$ and $Y$ are independent or not, and compute $\operatorname{Cov}(X,Y)$.

$$\rightarrow f_Y(y)=-\log(1-y)\mathbb{I}_{[0,1]}(y)\Rightarrow f_X(x)f_Y(y)\neq f_{XY}(x,y)\\
\Rightarrow X\text{ and }Y\text{ are not independent}$$
$$\rightarrow \operatorname{Cov}(X,Y)=-\frac{1}{6}$$

3) Now we assume to break the stick in the points $X$ and $Y$, and to form a triangle with the pieces that we have. Remembering that in a triangle the sum of the lengths of two sides must be greater than the length of the third side, what is the probability to form a triangle with the three pieces of the stick?


I'm stuck on point 3). How would you fix it?
Thanks in advance for any help.
 A: If the sum of the lengths of two sides must be greater than the third side, that means that each side cannot be greater than $0.5$ so the probability is
$$\mathbb{P}[Y-X<\frac{1}{2};X<\frac{1}{2};Y>\frac{1}{2}]$$
Graphically:

In formula:
$$\int_0^{\frac{1}{2}} \frac{1}{1-x}dx\int_{\frac{1}{2}}^{x+\frac{1}{2}} dy=\frac{2ln2-1}{2}\approx 0.19$$
A: Let $S$ be the region in the $xy$-plane defined by the constraints
$$
\left\lbrace
\begin{align*}
0 < x < 1\\[4pt]
x\le y < 1\\[4pt]
\end{align*}
\right.
$$
Then the joint density function of the random variables $X,Y$ is given by
$$
f(x,y)=
\begin{cases}
{\Large{\frac{1}{1-x}}}&\text{if}\;\,(x,y)\in S\\[4pt]
0&\text{otherwise}\\
\end{cases}
$$
Then we get
\begin{align*}
E[X]&=\int_0^1\int_x^1 x\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{1}{2}\\[4pt]
E[Y]&=\int_0^1\int_x^1 y\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{3}{4}\\[4pt]
E[XY]&=\int_0^1\int_x^1 xy\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{5}{12}\\[4pt]
\end{align*}
hence $X,Y$ are not independent since
$$
E[X]{\,\cdot\,}E[Y]
=
\frac{1}{2}{\,\cdot\,}\frac{3}{4}
=
\frac{3}{8}
\ne
\frac{5}{12}
=
E[XY]
$$
For the covariance, we get
$$
\text{Cov}(X,Y)
=
\int_0^1\int_x^1 
\left(
\Bigl(x-\frac{1}{2}\Bigr)
\Bigl(y-\frac{3}{4}\Bigr)
\right)
\!
\Bigl(\frac{1}{1-x}\Bigr)
\;dy\;dx
=
\frac{1}{24}
$$
The potential triangle has side lengths $a,b,c$ where
$$
\left\lbrace
\begin{align*}
a&=x\\[4pt]
b&=y-x\\[4pt]
c&=1-y\\[4pt]
\end{align*}
\right.
$$
hence noting that $a+b+c=1$, the triangle inequalities are satisfied if and only if
$0 < a,b,c < {\large{\frac{1}{2}}}$.

Computing $P\bigl(a \ge {\large{\frac{1}{2}}}\bigr)$, we get
$$
P\Bigl(a\ge\frac{1}{2}\Bigr)
=
\int_{\large{\frac{1}{2}}}^1\int_x^1 \frac{1}{1-x}\;dy\;dx=\frac{1}{2}
$$
Computing $P\bigl(b\ge{\large{\frac{1}{2}}}\bigr)$, we get
$$
P\Bigl(b\ge\frac{1}{2}\Bigr)
=
\int_0^{\large{\frac{1}{2}}}\int_{{\large{x+{\large{\frac{1}{2}}}}}}^1 \frac{1}{1-x}\;dy\;dx
=
\frac{1}{2}-\frac{1}{2}\ln(2)
$$
Computing $P\bigl(c\ge{\large{\frac{1}{2}}}\bigr)$, we get
$$
P\Bigl(c\ge\frac{1}{2}\Bigr)
=
\int_0^{{\large{\frac{1}{2}}}}
\int
_
{\large{x}}
^
{{\large{\frac{1}{2}}}}
\frac{1}{1-x}\;dy\;dx
=
\frac{1}{2}-\frac{1}{2}\ln(2)
$$
In the region $S$, we have $0 < a,b,c < 1$, hence since $a+b+c=1$, at most one of $a,b,c$ can be at least ${\large{\frac{1}{2}}}$.

It follows that the probability that $a,b,c$ qualify as the side lengths of triangle is given by
\begin{align*}
&
1
-
\left(
P\Bigl(a\ge\frac{1}{2}\Bigr)
+
P\Bigl(b\ge\frac{1}{2}\Bigr)
+
P\Bigl(c\ge\frac{1}{2}\Bigr)
\right)
\\[4pt]
=&
1
-
\left(
\left(
\frac{1}{2}
\right)
+
\left(
\frac{1}{2}-\frac{1}{2}\ln(2)
\right)
+
\left(
\frac{1}{2}-\frac{1}{2}\ln(2)
\right)
\right)
\\[4pt]
=&
-\frac{1}{2}
+
\ln(2)
\approx .193
\\[4pt]
\end{align*}
