$ABCD$ is circumscribed in a circle $k$, such that $[ACB]=s$, $[ACD]=t$, $s<t$. Determine the smallest value of $\frac{4s^2+t^2}{5st}$ and when this minimum is achieved.
1. **Given**: \(ABCD\) is a quadrilateral circumscribed in a circle \(k\), with \([ACB] = s\) and \([ACD] = t\), where \(s < t\). We need to determine the smallest value of \(\frac{4s^2 + t^2}{5st}\).

2. **Inequality Analysis**: We start by analyzing the expression \(\frac{4s^2 + t^2}{5st}\). To find the minimum value, we use the inequality \(4s^2 + t^2 \geq 4st\). This inequality can be derived from the fact that the sum of squares is always greater than or equal to twice the product of the terms, i.e., \(a^2 + b^2 \geq 2ab\).

3. **Applying the Inequality**:
   \[
   4s^2 + t^2 \geq 4st
   \]
   Dividing both sides by \(5st\):
   \[
   \frac{4s^2 + t^2}{5st} \geq \frac{4st}{5st} = \frac{4}{5}
   \]

4. **Example to Achieve Minimum**: To show that the minimum value of \(\frac{4}{5}\) can be achieved, consider a specific configuration of the quadrilateral \(ABCD\). Let \(|AB| = |BC| = 1\) and \(|CD| = |DA| = \sqrt{2}\), with \(\angle ADC = \angle ABC = \frac{\pi}{2}\).

5. **Verification Using Ptolemy's Theorem**: Ptolemy's theorem states that for a cyclic quadrilateral \(ABCD\), the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals:
   \[
   |AB| \cdot |CD| + |BC| \cdot |DA| = |AC| \cdot |BD|
   \]
   Substituting the given lengths:
   \[
   1 \cdot \sqrt{2} + 1 \cdot \sqrt{2} = |AC| \cdot |BD|
   \]
   \[
   2\sqrt{2} = |AC| \cdot |BD|
   \]
   Since \(|AC| = \sqrt{2}\) (diagonal of a square with side 1), we get:
   \[
   |BD| = \frac{2\sqrt{2}}{\sqrt{2}} = 2
   \]

6. **Angles and Areas**: Given \(\angle ADC = \angle ABC = \frac{\pi}{2}\), the areas \([ACB]\) and \([ACD]\) can be calculated using the right triangle properties. For \(\triangle ACB\):
   \[
   s = \frac{1}{2} \cdot |AB| \cdot |BC| \cdot \sin(\angle ABC) = \frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}
   \]
   For \(\triangle ACD\):
   \[
   t = \frac{1}{2} \cdot |AC| \cdot |CD| \cdot \sin(\angle ACD) = \frac{1}{2} \cdot \sqrt{2} \cdot \sqrt{2} \cdot 1 = 1
   \]

7. **Substituting into the Expression**:
   \[
   \frac{4s^2 + t^2}{5st} = \frac{4 \left(\frac{1}{2}\right)^2 + 1^2}{5 \cdot \frac{1}{2} \cdot 1} = \frac{4 \cdot \frac{1}{4} + 1}{\frac{5}{2}} = \frac{1 + 1}{\frac{5}{2}} = \frac{2}{\frac{5}{2}} = \frac{2 \cdot 2}{5} = \frac{4}{5}
   \]

Thus, the minimum value of \(\frac{4s^2 + t^2}{5st}\) is indeed \(\frac{4}{5}\), and it is achieved when \(|AB| = |BC| = 1\), \(|CD| = |DA| = \sqrt{2}\), and \(\angle ADC = \angle ABC = \frac{\pi}{2}\).

The final answer is \(\boxed{\frac{4}{5}}\).