Answer: 72

![](https://cdn.mathpix.com/cropped/2024_05_06_729073679a7806c16170g-2.jpg?height=537&width=766&top_left_y=1850&top_left_x=151)

In the image, there are three rectangles $2 \times (a+b)$ and three rectangles $2 \times (c+d)$, forming a $6 \times 6$ square, since $a+b+c+d=6$. The segment of length 2 in the center of the image is divided into segments $e$ and $f$, which sum to 2, as required by the condition.

The broken line depicted in the image, going from one corner of the square to the other, by the Pythagorean theorem, has a square of its length equal to the expression whose minimum value we need to find. Such a broken line can be constructed for each set of six numbers, so we only need to find the length of the shortest broken line, which is the diagonal of the square. The square of its length is $2 \cdot 6^{2}=$ 72.
# Problem 6. (4 points)

Six positive numbers, not exceeding 3, satisfy the equations $a+b+c+d=6$ and $e+f=2$. What is the smallest value that the expression

$$
\left(\sqrt{a^{2}+4}+\sqrt{b^{2}+e^{2}}+\sqrt{c^{2}+f^{2}}+\sqrt{d^{2}+4}\right)^{2}
$$

can take?