We are given the function:
[tex]\[ f(x) = x^2 - 4x + \left( \frac{a^2 + 6}{b+c} + \frac{b^2 + c}{a+c} + \frac{c^2 + a}{b+a} \right) \][/tex]
with the constraint [tex]\(a + b + c = 1\)[/tex], where [tex]\(a, b, c \in \mathbb{R}^+\)[/tex].
We need to find the minimum value of [tex]\(f(0)\)[/tex].
1. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4 \cdot 0 + \left( \frac{a^2 + 6}{b+c} + \frac{b^2 + c}{a+c} + \frac{c^2 + a}{b+a} \right) \][/tex]
[tex]\[ f(0) = \frac{a^2 + 6}{b+c} + \frac{b^2 + c}{a+c} + \frac{c^2 + a}{b+a} \][/tex]
2. Substitute the constraint [tex]\( a + b + c = 1 \)[/tex]:
Given [tex]\( a + b + c = 1 \)[/tex], we can express [tex]\( c \)[/tex] in terms of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ c = 1 - a - b \][/tex]
Substitute [tex]\(c\)[/tex] into the function:
[tex]\[ f(0) = \frac{a^2 + 6}{b + (1-a-b)} + \frac{b^2 + (1-a-b)}{a + (1-a-b)} + \frac{(1-a-b)^2 + a}{b + a} \][/tex]
[tex]\[ f(0) = \frac{a^2 + 6}{1 - a} + \frac{b^2 + 1 - a - b}{1 - b} + \frac{(1 - a - b)^2 + a}{b + a} \][/tex]
3. Finding the minimum value:
To minimize this function, we find the combination of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] that yields the smallest value for [tex]\( f(0) \)[/tex].
The numerical solution obtained previously tells us that the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] that nearly minimize [tex]\(f(0)\)[/tex]:
- [tex]\(a \approx 0\)[/tex]
- [tex]\(b \approx 0.5\)[/tex]
- [tex]\(c \approx 0.5\)[/tex]
Thus, substituting back these values, we focus on the minimum value which is approximated as:
[tex]\[ f(0) \approx 8.000000000025004 \][/tex]
4. Conclusion:
The integral part of the minimum value of [tex]\(f(0)\)[/tex] is [tex]\( \boxed{8} \)[/tex].
