$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.

Find the minimum value of $a+b+c$.

Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
Without loss of generality, suppose that $a\leq b\leq c$.  Also, write $N:=1001$.  Since $$(a+1)(b+1)(c+1)=N\,,$$ we have that $$|a+1|\,|b+1|\leq N\,.$$  This clearly means
$$|a+1|+|b+1|\leq N+1\,.$$
(If $x$ and $y$ are positive integers with $xy\leq N$, then $(x-1)(y-1)\geq 0$, which implies that $x+y\leq xy+1\leq N+1$.)
Thus,
$$-(a+1)-(b+1)\leq |a+1|+|b+1|\leq N+1\text{ or }a+b\geq -N-3\,.$$
Since $c+1$ is clearly positive, we have $c+1\geq 1$, which means $c\geq 0$.  That is,
$$a+b+c\geq -N-3\,.$$
The equality holds iff $(a,b,c)=(-N-1,-2,0)$.



If $a$, $b$, and $c$ are required to be nonnegative, then by the AM-GM Inequality, we have 
$$\frac{(a+1)+(b+1)+(c+1)}{3}\geq \sqrt[3]{(a+1)(b+1)(c+1)}=\sqrt[3]{1001}&gt;10\,.$$
That is, $a+b+c&gt;27$, or $a+b+c\geq 28$.  As $(a,b,c)=(6,10,12)$ works, the minimum value of $a+b+c$ is indeed $28$.  (Unlike the first part of this answer, $1001$ can't be replaced by an arbitrary positive integer $N$.  For each $N$, a different analysis is required.)
I am assuming here $a=b=c$ since $1000$ is the cube of $10$.

$$(a+1)(b+1)(c+1) = 1000$$

now, $(a+1)^3 = 1000$;
$$a+1 = 10\iff
a=9$$

so, $a+b+c = 3*9 = 27$. 

Just that we can get the integer value but I cannot say $27$ is the minimum value for $a+b+c$.