What is the probability that the largest of 3 independent uniformly distributed random variables is greater than the sum of the other two? I wanted to solve this using order statistics, therefore, I need to compute: 
$$\mathbb{P}\{X_{(1)}+X_{(2)} < X_{(3)}\}$$
The probability density function of $n$ order statistics is: 
$$n!f(x_1)\cdots f(x_n)$$
Therefore I need to solve the integral 
$$\iiint_{x_1+x_2 < x_3}3!\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3$$
Since $X_1,X_2,X_3$ are independent and uniformly distributed on $(0,1)$
To calculate the bounds, I do the following: 
$$\{0<x_1<1\ ,\ 0<x_2<1\ ,\ 0<x_3<1\ ,\ x_1+x_2<x_3\}$$
$$\{0<x_1<\min(1,x_3-x_2) \ ,\ 0<x_2<1\ ,\ 0<x_3<1\}$$
$$\{0<x_1<1, 1<x_3-x_2 \ ,\ 0<x_2<1\ ,\ 0<x_3<1\} \cup \{0<x_1<x_3-x_2, 1>x_3-x_2 \ ,\ 0<x_2<1\ ,\ 0<x_3<1\}$$
$$\{0<x_1<1  \ ,\ 0<x_2<\min(1,x_3-1)\ ,\ 0<x_3<1\} \cup \{0<x_1<x_3-x_2 \ ,\ \max(0, x_3-1)<x_2<1\ ,\ 0<x_3<1\}$$
Which reduces to
$$\{0<x_1<x_3-x_2 \ ,\ 0<x_2<1\ ,\ 0<x_3<1\}$$
This implies that
$$3!\int_0^1\int_0^1\int_0^{x_3-x_2}\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3=0$$
What am I doing wrong? Can I even apply order statistics to this problem or not?
 A: The joint pdf for all three sufficient statistics $(X_{(1)},X_{(2)},X_{(3)})$ is
$$
f(x_1,x_2,x_3) = 3! \cdot \mathbb{1}_{[ 0 \le x_1 \le x_2 \le x_3 \le 1]}.
$$
This is one part you're missing, I think youre integrating incorrectly over regions where your pdf should give zero.
Then the desired probability is
$$
\int 3! \cdot \mathbb{1}_{[ 0 \le x_1 \le x_2 \le x_3 \le 1]} \cdot \mathbb{1}_{[ x_1 + x_2 \le x_3]} \; dx_3 \, dx_2 \, dx_1
$$
and it may be helpful to rewrite this as
$$
\int 3! \cdot \mathbb{1}_{[ 0 \le x_1 \le x_2 \le x_3 \le 1]} \cdot \mathbb{1}_{[0 \le  x_1 + x_2 \le x_3 \le 1]} \; dx_3 \, dx_2 \, dx_1
$$
to get the integral bounds. Can you take it from here? ( I can post more if needed)
A: 
Which reduces to 
  $\{0<x_1<x_3−x_2 , 0<x_2<1 , 0<x_3<1\}$

No, it reduces to $\{0<x_1<\min\{x_2,x_3-x_2\}, 0<x_2<\bbox[2pt,border:red 1pt dotted]{x_3}, 0<x_3<1\}$
PS: recalling that $\{x_1<x_2<x_3\}$
A: The joint density function of the order statistics $X_{(i)}$ and $X_{(j)}$ when $i<j$ is $$ f_{X_{(i)}, X_{(j)}}\left(x_{i}, x_{j}\right)=\frac{n !}{(n-j) !(j-i-1) !(i-1) !}\left[F\left(x_{i}\right)\right]^{i-1}\left[F\left(x_{j}\right)-F\left(x_{i}\right)\right]^{j-i-1}\left[1-F\left(x_{j}\right)\right]^{n-j} f\left(x_{i}\right) f\left(x_{j}\right) $$
for all $ 0<x_{(i)}<x_{(j)}<1 $.
So, we need to calculate integration:
$$
\mathbb{P}(X_1+X_2\leq X_3)=\int_0^{1/2}\int_{x_1}^{1-x_1}\int_{x_1+x_2}^1 (1-x_1-x_2)(1-x_2) dx_3dx_2dx_1 = 1/2
$$
