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[tex]$
\begin{array}{l}
f(x) = x^2 - 4x + \left( \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \right) \\
a + b + c = 1 \quad a, b, c \in \mathbb{R} \\
f(0) = \text{mlhimo}
\end{array}
$[/tex]

Determina el valor mínimo del rango de la función.


Sagot :

Alright, let's break down the given problem and solve it step-by-step to find the minimum integer value in the range of the function [tex]\( f(x) \)[/tex].

### Step 1: Understanding the Function

The given function is:
[tex]\[ f(x) = x^2 - 4x + \left( \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \right) \][/tex]

### Step 2: Simplifying the Constant Term

The constant term in the function is:
[tex]\[ \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \][/tex]
Given that [tex]\( a + b + c = 1 \)[/tex], we need to work on simplifying this term.

### Step 3: Finding the Minimum of the Quadratic Expression

The quadratic expression [tex]\( x^2 - 4x + C \)[/tex] is a standard quadratic function where [tex]\(C\)[/tex] is a constant derived from the problem. The minimum value of this quadratic expression can be found using the vertex formula for [tex]\(ax^2 + bx + c\)[/tex], where the vertex [tex]\(x\)[/tex] coordinate is given by:
[tex]\[ x = \frac{-b}{2a} \][/tex]
For our function:
[tex]\[ a = 1, \; b = -4 \][/tex]
So,
[tex]\[ x = \frac{-(-4)}{2 \cdot 1} = \frac{4}{2} = 2 \][/tex]

### Step 4: Evaluating the Function at the Vertex

Now, substituting [tex]\( x = 2 \)[/tex] back into the function:
[tex]\[ f(2) = 2^2 - 4 \cdot 2 + C \][/tex]
[tex]\[ f(2) = 4 - 8 + C \][/tex]
[tex]\[ f(2) = -4 + C \][/tex]

### Step 5: Finding the Constant Term

Now we need to determine the value of the constant term [tex]\( C \)[/tex]:
[tex]\[ C = \frac{a^2 + b}{b + c} + \frac{b^2 + c}{a + c} + \frac{c^2 + a}{b + a} \][/tex]
Let's use symmetry and the condition [tex]\( a + b + c = 1 \)[/tex] to simplify further. By trial of typical symmetrical values:

Let's assume [tex]\( a = b = c \)[/tex]
Since [tex]\( a + b + c = 1 \)[/tex], if [tex]\( a = b = c \)[/tex], then:
[tex]\[ 3a = 1 \Rightarrow a = \frac{1}{3} \][/tex]
Thus:
[tex]\[ C = \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} + \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} + \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{1}{3} + \frac{1}{3}} \][/tex]

Simplify each term individually:
[tex]\[ \frac{\left(\frac{1}{3}\right)^2 + \frac{1}{3}}{\frac{2}{3}} = \frac{\frac{1}{9} + \frac{1}{3}}{\frac{2}{3}} \][/tex]
[tex]\[ \frac{\frac{1}{9} + \frac{3}{9}}{\frac{2}{3}} = \frac{\frac{4}{9}}{\frac{2}{3}} = \frac{4}{9} \cdot \frac{3}{2} = \frac{4}{3 \cdot 3} \cdot 3 = \frac{2}{3} \][/tex]

Adding up the three terms:
[tex]\[ C = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = 2 \][/tex]

### Step 6: Evaluate the Minimum Value of the Function

Using [tex]\( C = 2 \)[/tex]:
[tex]\[ f(2) = -4 + 2 = -2 \][/tex]

### Conclusion: Minimum Integer Value

The minimum value of the function [tex]\( f(x) \)[/tex] is [tex]\(-2\)[/tex], which is already an integer.

Thus, the minimum integer value in the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{-2} \][/tex]