Expected distance from a random point chosen in a tetrahedron to its closest face 
A point is chosen at random in a regular tetrahedron of height 1(so that the probability of it lying in any given region is directly proportional to the volume of that region). Find the expected value of the distance from the point to its closest face.

I have solved this question using a geometric approach that takes 6 pages of arduous mathematics, however after looking at the provided solution, they seem to have a slick approach that I just can't see the rationale behind.
Let $x_1,x_2,x_3,x_4$ be perpendicular distances from P to each face respectively. Let $X=\min \{x_1,x_2,x_3,x_4\}$. Then $x_1+x_2+x_3+x_4=1$ so $P(X\ge x)\propto (\frac{1}{4}-x)^3$ so the pdf of $X$ is $f(x)=k(\frac{1}{4}-x)^2$ and $E(X)=\frac{1}{16}$ follows.
($x_1+x_2+x_3+x_4=1$ is the geometric fact that they sum to the height of the tetrahedron.). I do not understand the $P(X\le x)\propto (\frac{1}{4}-x)^3$ part.
 A: The given hint is misleading (if you type it correctly) and the correct expression should be:
$$
P(X\le x)\propto \left(\tfrac{1}{4}\right)^3-\left(\tfrac{1}{4}-x\right)^3\quad\text{or}\quad P(X\ge x)\propto\left(\tfrac{1}{4}-x\right)^3
$$
provided that $x\le\tfrac14$. Indeed the original form implies obviously wrong conclusion that the larger is $x$ the smaller is the probability that the least distance does not exceed $x$. Besides the pdf $f(x)$ which is the derivative of $P(X\le x)$ appears to be negative.
Besides the geometrical way which was indicated in the comment one can compute the pdf in pure algebraic way. For this consider the equations:
$$
x_1,x_2,x_3\ge x,\quad x_1+x_2+x_3=1-x.\tag1
$$
Due to the last equation one of the variables ($x_3$) is uniquely determined by the other three and can be ignored provided that it satisfies the condition $x_3\ge x$. Accordingly the probability density that the value $x$ represents the least distance to a tetrahedron face is proportional to the volume of the configuration space left for $x_1,x_2$ to satisfy the conditions $(1)$.
It is easy to check that when $x$ is fixed $x_1$ has to satisfy the inequality $x_1\le1-3x$. Indeed
$$
1-x=x_1+x_2+x_3\ge x_1+2x\implies 1-3x\ge x_1.
$$
Similarly one concludes that when $x$ and $x_1$ are fixed,
$x_2$ has to satisfy $x_2\le1-2x-x_1$.
With the established boundary the volume of the configuration space can be readily computed as:
$$
V(x)=\int_x^{1-3x}dx_1\int_x^{1-2x-x_1}dx_2=8\left(\tfrac14-x\right)^2,
\quad 0\le x\le\tfrac14,
$$
and the corresponding probability density function is:
$$
f(x)=\frac{V(x)}{\int_0^{1/4}V(x)dx}=192\left(\tfrac14-x\right)^2.
$$
