We use the following two tools:
- Tool 1: Every real number $x$ satisfies $\lfloor x\rfloor>x-1$ (with strict inequality).
- Tool 2: All positive real numbers $y$ and $z$ satisfy
$\frac{x}{y}+\frac{y}{x}\ge2$; indeed this inequality is equivalent to $y^2+z^2\ge2yz$ and $(y-z)^2\ge0$.
Now the expression $E$ that is to be minimized can be estimated (by using these two tools) as follows:
\begin{eqnarray*}
E &=&
\left\lfloor\frac{b+c+d}{a} \right\rfloor+
\left\lfloor\frac{a+c+d}{b} \right\rfloor+
\left\lfloor\frac{a+b+d}{c} \right\rfloor+
\left\lfloor\frac{a+b+c}{d} \right\rfloor
\\ &>&
\left(\frac{b+c+d}{a}-1\right)+
\left(\frac{a+c+d}{b}-1\right) +
\left(\frac{a+b+d}{c}-1\right) +
\left(\frac{a+b+c}{d}-1\right)
\\ &=&
\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
\\ &\ge& ~6\cdot2-4 ~~=~~ 8.
\end{eqnarray*}
We get $E>8$. Since $E$ is the sum of four integers and hence integer itself, this actually implies the lower bound $E\ge9$.
Finally, for $a=4$ and $b=c=d=5$ the expression attains the value $9$.
Hence the smallest possible value of the above expression is
$~~~9~$ (nine)
For the Bonus question, the above argument easily generalizes. The resulting lower bound is
$$ E ~>~ {n\choose 2}\cdot 2 -n, $$
which implies $E\ge(n-1)^2$.
By setting $a_1=n$ and $a_2=a_3=\cdots=a_n=n+1$, we get
$$\left\lfloor\frac{a_2+a_3+\cdots+a_n}{a_1} \right\rfloor =
\left\lfloor\frac{(n-1)(n+1)}{n} \right\rfloor =
\left\lfloor n-\frac{1}{n} \right\rfloor = n-1,$$
and
$$\left\lfloor\frac{a_1+a_3+\cdots+a_n}{a_2} \right\rfloor =
\left\lfloor\frac{(n-2)(n+1)+n}{n+1} \right\rfloor =
\left\lfloor (n-2)+\frac{n}{n+1} \right\rfloor = n-2.$$
The resulting value of the expression is $(n-1)+(n-1)(n-2)=(n-1)^2$, and hence coincides with the lower bound.
Summarizing, the smallest possible value of the expression in the Bonus question
$~~~(n-1)^2$
