Let $x,~y$ and $1 - x - y$ denote the lengths of the broken pieces, where $x, y \sim U(0,1)$ and $x + y \le 1$. Let us define $z \dot{=} \min(x,y,1-x-y)$. We are interested in finding the probability: $P\left(z \le {1\over 5} \right) $.
We will compute $P\left(z \le z_0 \right)$ as $1 - P\left(z > z_0 \right)$.
\begin{equation}
\tag{1}
P\left(z > z_0 \right) = {\int_0^1 dx \int_0^{1-x} dy~\Theta(x - z_0).~\Theta(y - z_0).~\Theta(1 - x - y -z_0) \over \int_0^1 dx \int_0^{1-x} dy }
\end{equation}
where $\Theta(a) \equiv \mathbb{1}_{a>0} = 1$ if $(a > 0),~0$ otherwise. Note that $ x + y \le 1$ and the probability of this happening is given by the integral in the denominator which evaluates to ${1\over 2}$. The above expression for the probability easily generalizes to other distributions (defined on the unit interval) and can also be extended to have more variables. The probability in equation (1) simplifies to:
\begin{equation}
P\left(z > z_0 \right) = 2 {\int_0^1 dx \int_0^{1-x} dy~\Theta(x - z_0).~\Theta(y - z_0).~\Theta(1 - x - y -z_0)}
\end{equation}
The integral in the above equation is essentially the area of the triangle ABC shown in the figure (below).
Integration Domain (depicted by shaded triangle ABC)
The integral can also be evaluated directly by finding the limits of $x$ and $y$ integrals to get rid of the step functions. The limits of the $x$ and $y$ integral can be obtained by solving for the coordinates of $A$, $B$ and $C$ (pairwise intersections of lines $P$, $Q$ and $R$). (Aside: There might be other variants of the problem where $P$, $Q$ and $R$ are more complicated curves and not simple lines. In such cases one needs to evaluate the integral within the area bounded by the curves $P,~Q$ and $R$. In the present case, the integral simplifies to:
$$
P\left(z > z_0 \right) = 2 \int_{z_0}^{1-2 z_0} dx \int_{z_0}^{1 - x - z_0} dy = (1 - 3 z_0)^2 = 2 \Delta ABC \qquad \text{for } ~z_0 < {1\over 3}
$$
Note that when $z_0$ is greater than one-third, the probability is trivially zero. This can also be inferred from the fact that the lower limit of the $x$ integral exceeds the upper limit and hence one of the step functions must evaluate to zero in this case.
Returning to the original problem:
$$
P\left(z \le z_0 \right) = 1 - (1 - 3 z_0)^2 \qquad \text{for } ~z_0 < {1\over 3}
$$
With $z_0 = {1\over 5}$, we get:
$$
P\left(\text{min length of the three pieces} \le {1\over 5} \right) = {21\over 25}
$$
The problem can be generalized to $n$ pieces and to other distributions on the unit interval. The above procedure should work for such variants.
