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

Question

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.

Answer 1

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 $$\frac{(a+1)+(b+1)+(c+1)}{3}\geq \sqrt[3]{(a+1)(b+1)(c+1)}=\sqrt[3]{1001}>10\,.$$ That is, $a+b+c>27$, 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.)

Answer 2

Hint:

Expand $$(a+1)(b+1)(c+1)$$

And notice $$1001=7\cdot11\cdot13$$

Answer 3

Trial and error can be avoided. Note that minimizing $x+y+z$ where the positive real numbers $x,y,z$ have a specified product $1001$ is achieved by making $x,y,z$ as nearly equal as possible.

Here we minimize $(1+a)+(1+b)+(1+c)$ when we minimize $a+b+c$. Since $a,b,c$ are required to be (positive?) integers, the most equal we can make the factors is by assigning the primes $7,11,13$ to $(1+a),(1+b),(1+c)$ in some order. Although this assignment is not unique, the minimum sum $a+b+c = 6+10+12$ will be unique.

Answer 4

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 $a+b+c$.