In the tetrahedron $ABCD$, $E$ and $F$ are the midpoints of $BC$ and $AD$, $G$ is the midpoint of the segment $EF$. Construct a plane through $G$ intersecting the segments $AB$, $AC$, $AD$ in the points $M,N,P$ respectively in such a way that the sum of the volumes of the tetrahedrons $BMNP$, $CMNP$ and $DMNP$ to be minimal.

[i]H. Lesov[/i]
1. **Define the problem and variables:**
   - We are given a tetrahedron \(ABCD\) with \(E\) and \(F\) as the midpoints of \(BC\) and \(AD\) respectively.
   - \(G\) is the midpoint of segment \(EF\).
   - We need to construct a plane through \(G\) intersecting segments \(AB\), \(AC\), and \(AD\) at points \(M\), \(N\), and \(P\) respectively, such that the sum of the volumes of tetrahedrons \(BMNP\), \(CMNP\), and \(DMNP\) is minimized.

2. **Introduce ratios and constraints:**
   - Let \(b = \frac{AB}{AM}\), \(c = \frac{AC}{AN}\), and \(d = \frac{AD}{AP}\).
   - Note that \(b + c + d = 4\). This follows from the vector equation \(\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} = 4\overrightarrow{AG}\) and the fact that \(\beta\overrightarrow{AM} + \gamma\overrightarrow{AN} + \delta\overrightarrow{AP} = \overrightarrow{AG}\) for some nonnegative \(\beta, \gamma, \delta\) with \(\beta + \gamma + \delta = 1\).

3. **Volume calculation:**
   - The volume of tetrahedron \(BMNP\) can be expressed as a fraction of the volume of \(ABCD\). Specifically, \([BMNP] = \frac{1}{bcd}[ABCD]\).
   - Similarly, the volumes of \(CMNP\) and \(DMNP\) can be expressed as \(\frac{1}{bcd}[ABCD]\).

4. **Sum of volumes:**
   - The sum of the volumes of tetrahedrons \(BMNP\), \(CMNP\), and \(DMNP\) is:
     \[
     \sum_{cyc}[BMNP] = \sum_{cyc}\left(\frac{1}{bcd}[ABCD]\right)
     \]
   - This simplifies to:
     \[
     [ABCD] \sum_{cyc} \frac{1}{bcd}
     \]

5. **Minimization using AM-GM inequality:**
   - By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
     \[
     \frac{b + c + d}{3} \geq \sqrt[3]{bcd}
     \]
   - Given \(b + c + d = 4\), we get:
     \[
     \frac{4}{3} \geq \sqrt[3]{bcd} \implies \left(\frac{4}{3}\right)^3 \geq bcd \implies \frac{64}{27} \geq bcd
     \]
   - Therefore:
     \[
     \frac{1}{bcd} \geq \frac{27}{64}
     \]

6. **Conclusion:**
   - The minimum value of \(\frac{1}{bcd}\) is \(\frac{27}{64}\), which is achieved when \(b = c = d = \frac{4}{3}\).
   - Thus, the sum of the volumes of tetrahedrons \(BMNP\), \(CMNP\), and \(DMNP\) is minimized when \(MNP\) is parallel to \(BCD\).

The final answer is \( \boxed{ \frac{27}{64}[ABCD] } \).