Multiple integral of min How can this be proven?
$$
\int_0^1 \int_0^1...\int_0^1 min(x_1,x_2,...,x_n) dx_1dx_2...dx_n = \frac1{n+1} $$
I tried to split the last integral in $$\int_0^{min(x_1,x_2,...,x_{n-1})} x_ndx_n + \int_{min(x_1,x_2,...,x_{n-1})}^1 min(x_1,x_2,...,x_{n-1})dx_n$$
And then continue in this manner. 
So the function that remains to be integrated satisfies this recurrence relation
$$ f_n = (1-x)f_{n-1} + \int f_{n-1}  $$
I then thought to use the following notation 
$$I_n = \frac{ (-1)^{n+1}x^n }{n} +  \frac{ (-1)^{n}x^{n-1} }{n-1} $$
If we compute the last integral, the n-1 integrals apply to $I_2$
If we apply another integral we get $I_3+I_2$ and so on, which will look like Pascal's triangle. But I can't go any further.
 A: Here is another opproach:
First consider the set $A=\{(t_1,\dots,t_n):t_1\le\dots\le t_n\}$. In this set, your integral becomes
$$
\int_0^1 \int_0^{t_n}...\int_0^{t_2} \int_0^{t_1} x_1 dx_1dx_2...dx_n = \frac1{(n+1)!}.
$$
Since there are $n!$ sets like $A$ which are the union is $[0,1]^n$ then 
$$
\int_0^1 \int_0^1...\int_0^1 min(x_1,x_2,...,x_n) dx_1dx_2...dx_n = n!\times  \frac1{(n+1)!}=\frac1{n+1}.
$$ 
A: First, write
\begin{align*}
I
&:= \int_{0}^{1} \cdots \int_{0}^{1} \min \{ x_{1}, \cdots, x_{n} \} \, dx_{n}\cdots dx_{1} \\
&= \int_{0}^{1} \cdots \int_{0}^{1} \int_{0}^{\min \{ x_{1}, \cdots, x_{n} \}} dt \, dx_{n}\cdots dx_{1}.
\end{align*}
By noting that $t \leq \min \{x_{1}, \cdots, x_{n}\}$ is equivalent to $t \leq x_{1}, \cdots, t \leq x_{n}$, we find that $I$ is the volume of the region $\mathcal{W}$ defined by
$$ \mathcal{W} = \{ (t, x_{1}, \cdots, x_{n}) : 0 \leq t \leq 1, t \leq x_{1} \leq 1, \cdots, t \leq x_{n} \leq 1 \}. $$
Thus by Fubini's Theorem we have
\begin{align*}
I
&= \int_{0}^{1} \int_{t}^{1} \cdots \int_{t}^{1} \, dx_{n}\cdots dx_{1} \, dt
 = \int_{0}^{1} (1-t)^{n} \, dt
 = \frac{1}{n+1}.
\end{align*}
This solution can be encoded in the language of probability as follows: Let $U_{1}, \cdots, U_{n}$ be i.i.d. uniform random variables on $[0, 1]$. Then
\begin{align*}
&\int_{0}^{1} \cdots \int_{0}^{1} \min \{ x_{1}, \cdots, x_{n} \} \, dx_{n}\cdots dx_{1} \\
&\qquad = \Bbb{E}( \min\{ U_{1}, \cdots, U_{n} \} ) \\
&\qquad = \int_{0}^{1} \Bbb{P}(\min\{ U_{1}, \cdots, U_{n} \} > x) \, dx \\
&\qquad = \int_{0}^{1} \Bbb{P}( U_{1} > x, \cdots, U_{n} > x) \, dx \\
&\qquad = \int_{0}^{1} \Bbb{P}( U_{1} > x)^{n} \, dx \qquad (\because \text{i.i.d.}) \\
&\qquad = \int_{0}^{1} (1 - x)^{n} \, dx
= \frac{1}{n+1}.
\end{align*}
A: Use the fact that, $\min \{a,b\} = (a+b-|a-b|)/2$ for all $a,b \in \mathbb R$.
