Computing $\mathbb{P}\{X_{(1)}+X_{(2)} > X_{(3)}\}$ where $X_1,X_2,X_3$ are i.i.d $U(0,1)$ 
Suppose that $X_1 , X_2 , X_3$ are independent $U (0, 1)$-distributed random variables and let $(X_{(1)} , X_{(2)} , X _{(3)} )$ be the corresponding order statistic. Compute $\mathbb{P}\{X_{(1)}+X_{(2)} > X_{(3)}\}$.

I have the joint of the distribution which is $3!$
Also, I have:   
$$\mathbb{P}\{X_{(1)}+X_{(2)} > X_{(3)}\}=1-\mathbb{P}\{X_{(1)}+X_{(2)} \le X_{(3)}\}$$ with $\mathbb 0<x_1<\min\{x_2,x_3-x_2\}, \quad 0<x_2<x_3, \quad0<x_3<1$ 
But when I do: 
$$1-6\left(\int_0^{1}\int_0^{x_3}\int_0^{x_2} \,dx_1\,dx_2\,dx_3 +\int_0^{1}\int_0^{x_3}\int_0^{x_3-x_2} \,dx_1\,dx_2\,dx_3\right)=-3$$ 
I do not really know what I'm doing wrong.
 A: There could possibly be a simpler argument that does not require doing the integration explicitly.
For the integration, I felt it convenient to use a change of variables.
You are looking for
$$P(X_{(1)}+X_{(2)}<X_{(3)})=6\iiint\mathbf1_{x+y<z\,,\,0<x<y<z<1}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$
Change variables $(x,y,z)\to(u,v,w)$ with $$u=x+y\,,\,v=y\,,\,w=z$$
Then, $$x+y<z\,,\,0<x<y<z<1\implies u<w\,,\,0<u-v<v<w<1$$
Therefore, 
\begin{align}
\iiint\mathbf1_{x+y<z\,,\,0<x<y<z<1}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z&=\int_0^1\int_0^w\int_v^{\min(2v,w)}\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w
\\&=\int_0^1\int_0^{w/2}\int_v^{2v}\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w+\int_0^1\int_{w/2}^w\int_v^w\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w
\\&=\frac{1}{12}
\end{align}
A: There’s a symmetry among the intervals that the $X_i$ divide the unit interval into. You can see this by imagining that you uniformly randomly choose $4$ points on a circle, then uniformly randomly choose one of the $4$ points at which to cut the circle into an interval, with the remaining $3$ points yielding the $X_i$ uniformly randomly independently distributed on the unit interval. Since the $4$ resulting intervals were on an equal footing in the circle, they are still so in the interval.
Denoting these intervals by $a$, $b$, $c$ and $d$ in order, we obtain $X_{(1)}+X_{(2)}=a+a+b$ and $X_{(3)}=a+b+c$, so $X_{(1)}+X_{(2)}\gt X_{(3)}$ is equivalent to $a\gt c$. Due to the symmetry among the intervals, the probability for this is $\frac12$.
