8. II 2 (NET 3) $\mathrm{IMO5}$ If $a, b, c, d$ are arbitrary positive real numbers, find all possible values of
$$
S=\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d} .
$$
8. It is clear that
$$
\begin{array}{c}
\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}<S \\
\text { and } S<\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d} .
\end{array}
$$
or equivalently, $1<S<2$.
On the other hand, all values from $(1,2)$ are attained. Since $S=1$ for $(a, b, c, d)=(0,0,1,1)$ and $S=2$ for $(a, b, c, d)=(0,1,0,1)$, due to continuity all the values from $(1,2)$ are obtained, for example, for $(a, b, c, d)=(x(1-x), x, 1-x, 1)$, where $x$ goes through $(0,1)$.
Serond sodution. Set
$$
S_{1}=\frac{a}{a+b+d}+\frac{c}{b+c+d} \quad \text { and } \quad S_{2}=\frac{b}{a+b+c}+\frac{d}{a+c+d} .
$$

We may assume without loss of generality that $a+b+c+d=1$. Putting $a+c=x$ and $b+d=y$ (then $x+y=1)$, we obtain that the set of values of
$$
S_{1}=\frac{a}{1-c}+\frac{c}{1-a}=\frac{2 a c+x-x^{2}}{a c+1-x}
$$
is $\left(x, \frac{2 x}{2-x}\right)$. Having the analogous result for $S_{2}$ in mind, we conclude that the values that $S=S_{1}+S_{2}$ can take are $\left(x+y, \frac{2 x}{2-x}+\frac{2 y}{2-y}\right)$. Since $x+y=1$ and
$$
\frac{2 x}{2-x}+\frac{2 y}{2-y}=\frac{4-4 x y}{2+x y} \leq 2
$$
with equality for $x y=0$, the desired set of values for $S$ is (1.2).