. Find the values that the quantity

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

can take when $a, b, c, d$ range over the strictly positive real numbers.
. On a

$$
\begin{aligned}
& \frac{a}{a+b+d}+\frac{b}{b+c+a}+\frac{c}{c+d+b}+\frac{d}{d+a+c} \\
& >\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1
\end{aligned}
$$

and also

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

Finally, by fixing $a$ and $b$ and letting $c$ and $d$ tend to 0, the quantity tends to 1. Similarly, by fixing $a$ and $c$ and letting the other two tend to 0, the quantity tends to 2. In conclusion, by the Intermediate Value Theorem, the proposed quantity describes the interval $] 1,2[$.