Uniformly at random, break a unit stick in two places. What is the probability that the smallest piece is $\leq 1/5$?

Dividing the unit stick in 3 pieces at random, we have

It is eviedent that the three sticks can be identified as follows:

1. $U=min(X,Y)$

2. $V=1-Max(X,Y)$

3. $Z=|X-Y|$

The probability that the minimum is Greater than $\frac{1}{5}$ is the probability that all 3 sticks are greater than $\frac{1}{5}$

Say

$$\mathbb{P}\Bigg[min(U,V,Z)>\frac{1}{5}\Bigg]=\mathbb{P}\Bigg[U>\frac{1}{5},V>\frac{1}{5},Z>\frac{1}{5}\Bigg]$$

Now given that $V>\frac{1}{5}$ is equivalent to $max(X,Y)<\frac{4}{5}$ the area results to me $$(0.8-0.4)^2=0.16$$

Which is the resulting intersection of the following 3 areas

Thus the requested probability is its complement to 1

$$ \bbox[5px,border:2px solid red] { \mathbb{P}\Bigg[min(U,V,Z)\leq\frac{1}{5}\Bigg]=1-0.16=0.84 \ } $$

Hint: Let $X$ and $Y$ be the coordinates of a point in the unit square. Draw a picture and shade the region where one piece is less than $\frac 15$. You should have the outer border of the square plus a region following the main diagonal. Evaluate the area of the shaded region. The region where the smallest piece is less than $\frac 15$ is the same as the region where at least one piece is less than $\frac 15$.