$$
\begin{array}{l}
\text { Three. (50 points) Given } a, b, c \in \mathbf{R}_{+}, a b c=1 \text {. Find } \\
f(a, b, c)=\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c a}+\frac{1}{a^{2}+a b}
\end{array}
$$

the minimum value.
Four. (50 points) Prove: There exists a real number set $X$ with no less than four elements, such that for any four distinct elements $a, b, c, d$ in $X$, there is $a b+c d \in X$, and prove that the elements in such a number set $X$ cannot all be positive.
$$
Three, let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$. Then
$\frac{1}{b^{2}+b c}=\frac{1}{\left(\frac{z}{y}\right)^{2}+\frac{z}{y} \cdot \frac{x}{z}}=\frac{y^{2}}{z^{2}+x y}$,
$\frac{1}{c^{2}+c a}=\frac{z^{2}}{x^{2}+y z}, \frac{1}{a^{2}+a b}=\frac{x^{2}}{y^{2}+z x}$.
Thus, $\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c a}+\frac{1}{a^{2}+a b}$
$=\frac{y^{2}}{z^{2}+x y}+\frac{z^{2}}{x^{2}+y z}+\frac{x^{2}}{y^{2}+z x}$
$=\frac{y^{4}}{y^{2} z^{2}+y^{3} x}+\frac{z^{4}}{z^{2} x^{2}+z^{3} y}+\frac{x^{4}}{x^{2} y^{2}+x^{3} z}$
$\geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{z^{3} y+y^{3} x+x^{3} z+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}$.
First, $y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} \leqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{3}$,
and by Vasc's inequality
$x^{3} z+y^{3} x+z^{3} y \leqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{3}$,

we get
$\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c a}+\frac{1}{a^{2}+a b}$
$\geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{x^{3} z+y^{3} x+z^{3} y+y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}}$
$\geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{\frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{3}+\frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{3}}=\frac{3}{2}$.
Equality holds if and only if $a=b=c=1$.
Therefore, $f_{\text {min }}(a, b, c)=\frac{3}{2}$.
Four, first for $X_{0}=\left\{0, \frac{1}{2}, 1,2\right\}$ we have
$0 \times \frac{1}{2}+1 \times 2=2 \in X_{0}$,
$0 \times 1+\frac{1}{2} \times 2=1 \in X_{0}$,
$0 \times 2+\frac{1}{2} \times 1=\frac{1}{2} \in X_{0}$,

Thus, there exists a set $X$ with at least four elements, such that for any four distinct elements $a, b, c, d$ in $X$, $a b+c d \in X$.

Next, we prove that the elements of set $X$ cannot all be positive.
Proof by contradiction.
Assume that all elements of set $X$ are positive. Then set $X$ can have at most one element greater than 1.

Otherwise, let $a, b, c, d$ be the four largest elements of set $X$, then $a>b>1>c>d$, thus, $a b+c d \in X$. But $a b+c d>a b>a$, which contradicts the fact that $a$ is the largest element of $X$.
Next, we prove that it is impossible for
$a>1 \geqslant b>c>d>0$

to form a four-element set $X=\{a, b, c, d\}$.
Suppose otherwise, then
$a b+c d>a c+b d>a d+b c>a d>d$,

Thus, it can only be
$a b+c d=a, a c+b d=b, a d+b c=c$.
Since $0<b)$. Since there exist $c, d \in X$ and $c \leqslant 1, d \leqslant 1$ such that $a, b, c, d$ are all distinct, then
$$
\begin{array}{l}
(a c+b d)-(a d+b c)=(a-c)(c-d) \neq 0, \\
|(a c+b d)-(a d+b c)|=(a-b)|c-d| \\
<a-b,
\end{array}
$$

which contradicts the minimality of $a-b$.
Thus, the assumption is incorrect, i.e., the elements of set $X$ that satisfy the requirement cannot all be positive.