Find the smallest possible value of the expression

$$
\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{b+c+d}{a}\right\rfloor+\left\lfloor\frac{c+d+a}{b}\right\rfloor+\left\lfloor\frac{d+a+b}{c}\right\rfloor,
$$

where \(a, b, c\) and \(d\) vary over the set of positive integers.

(Here \(\lfloor x\rfloor\) denotes the greatest integer less than or equal to \(x\).)
The answer is 9.

Notice that $\lfloor x\rfloor>x-1$ for all $x \in \mathbb{R}$. Therefore the given expression is strictly greater than

$$
\frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}-4,
$$

which can be rewritten as

$$
\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-4.
$$

Since $t+\frac{1}{t} \geq 2$ for $t>0$, we get that $6 \cdot 2-4=8$ is a strict lower bound for the given expression; since it takes integral values only, we actually get that 9 is a lower bound.

It remains to check that 9 can be attained; this happens for $a=b=c=5$ and $d=4$.