(BXMO 2014) Let $a, b, c$ and $d$ be strictly positive integers. Determine the smallest value that the expression:

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

can take.
With integer parts, there is not much to do: we apply the inequality $\lfloor x\rfloor>x-1$ to find, by rearranging the terms

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

and each term in parentheses is bounded below by 2 by AM-GM. Therefore, we have $S>8$, which means $S \geqslant 9$. By trial and error, we find the construction $(5,5,5,4)$.