Determine the range of values for

$$
S=\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}
$$

where $a, b, c,$ and $d$ are arbitrary positive numbers.
Given that $a, b, c, d$ are positive,

$$
S>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1
$$

and

$$
S<\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d}=2
$$

thus $1<S<2$, meaning the value set falls within the interval $(1,2)$. We will show that the value set fills the interval $(1,2)$.

First, let $a=b=x, c=d=1$. Then

$$
S=S(x)=\frac{2 x}{2 x+1}+\frac{2}{2+x}=\frac{2}{2+\frac{1}{x}}+\frac{2}{2+x}
$$

and here $x$ can be any positive number. When $x \rightarrow 0$, $S(x) \rightarrow 1$, and when $x=1$, $S(1)=\frac{4}{3}$. By the Bolzano theorem, the continuous function $S(x)$ takes on every value between 1 and $\frac{4}{3}$ in the interval $(0,1)$.

Second, let $a=c=x, b=d=1$. Then

$$
S=S(x)=\frac{2 x}{x+2}+\frac{2}{2 x+1}=\frac{2}{1+\frac{2}{x}}+\frac{2}{1+2 x}
$$

and $x$ can again be any positive number. When $x \rightarrow 0$, $S(x) \rightarrow 2$, and when $x=1$, $S(1)=\frac{4}{3}$. Again, by the Bolzano theorem, $S(x)$ takes on every value between $\frac{4}{3}$ and 2 in the interval $(0,1)$. Therefore, the value of $S$ runs through all numbers between 1 and 2.

Andor Hidjapusztai (Szeged, Radnóti M. Gymnasium, 4th grade)

Note. Without referring to limits, continuity, and the Bolzano theorem, we can also show that $S$ takes on every value $y$ between 1 and 2. For this, when $1<y \leq \frac{4}{3}$, we need to solve the equation

$$
\frac{2}{2+\frac{1}{x}}+\frac{2}{2+x}=y
$$

and when $\frac{4}{3} \leq y<2$, we need to solve the equation

$$
\frac{2}{1+\frac{2}{x}}+\frac{2}{1+2 x}=y
$$

Rearranging these, we get quadratic equations in $x$, whose solvability is guaranteed by the constraints on $y$. It can also be seen from these that the (existing) roots are always positive.

Erzsébet Lugosi (Cegléd, Kossuth L. Gymnasium, 3rd grade)