$a, b, c$ are integers and $a,b,c\geq 4$ If $4abc=(a+3)(b+3)(c+3)$ and $a, b, c$ are all not equal, then what is the value of $a+b+c$.
I tried solving by taking $a\neq b=c$ and couldn't find any integral value which could satisfy the condition.
How to proceed in the problem??
Suppose $a,b,c$ are integers, with $a,b,c \ge 4$, such that $$4abc=(a+3)(b+3)(c+3)$$
Assume $a\le b \le c$.
Suppose $b \ge 6$.
Then also $c\ge 6$, so $c+3 \le \bigl({\large{\frac{3}{2}}}\bigr)c$, hence \begin{align*} &4abc\le(a+3)(b+3)\bigl({\small{\frac{3}{2}}}\bigr)c\\[4pt] \implies\;&8ab\le 3(a+3)(b+3)\\[4pt] \implies\;&5ab\le 9a + 9b + 27\\[4pt] \implies\;&b(5a-9)\le 9a+27&&(*\!)\\[4pt] \implies\;&a(5a-9) \le 9a+27\\[4pt] \implies\;&5a^2-18a-27\le 0\\[4pt] \implies\;&a < 5\\[4pt] \implies\;&a=4\\[4pt] \implies\;&11b\le 63&&\text{[by substituting $a=4$ into$\;(*\!)$]}\\[4pt] \implies\;&b < 6\\[4pt] \end{align*} contradiction.
It follows that $b\le 5$.
Consider cases . . .
If $(a,b) = (4,4)$, then subsituting $a=4,b=4$ in the main equation yields $c={\large{\frac{49}{5}}}$, contradiction.
If $(a,b) = (4,5)$, then subsituting $a=4,b=5$ in the main equation yields $c=7$, which gives the solution triple $(a,b,c)=(4,5,7)$.
If $(a,b) = (5,5)$, then subsituting $a=5,b=5$ in the main equation yields $c={\large{\frac{16}{3}}}$, contradiction.
Thus, up to a permutation of the variables, the only solution is $(a,b,c)=(4,5,7)$, so $a+b+c=16$.
By symmetry we can demand $a \lt b \lt c$. If $a,b,c$ get too large the left side will grow much faster than the right. We can write your equation as $$3abc-3(ab+ac+bc)-9(a+b+c)-27=0\\abc-(ab+ac+bc)-3(a+b+c)-9=0$$ If $a=5$ this becomes $$4bc-8(b+c)-24=0\\(b-2)(c-2)-10=0$$ which has no solution with $b \ge 6, c\gt b$. It seems clear that greater $a$s will make things worse, so $a=4$. Then we have $$3bc-7(b+c)-21=0$$ and we note that the right side will have a factor $7$ from the $a+3$ so one of $b,c$ must be a multiple of $7$. We also note that $b+c$ must be a multiple of $3$. This screams to me $b=5,c=7$ and we can check that it works. I haven't really proved that there are no more solutions.
Without loss of generality, we may assume that $4\leq a \leq b \leq c,$ where the equalities does not hold for all.
Since $$4=\left(1+\frac{3}{a}\right)\left(1+\frac{3}{b}\right)\left(1+\frac{3}{c}\right)\leq \left(1+\frac{3}{a}\right)^3,$$ hence $$4\leq a\leq 5.$$
If $a=4.$ Then $$4=\left(1+\frac{3}{4}\right)\left(1+\frac{3}{b}\right)\left(1+\frac{3}{c}\right)\leq \left(1+\frac{3}{4}\right)\left(1+\frac{3}{b}\right)^2.$$ We have $$4=a\leq b \leq 5,$$which implies that the possible values of $b$ are $$b=4,5.$$ If $a=4,b=4$, then $c=\dfrac{49}{5},$ which is not an integer. If $a=4,b=5$, then $c=7$, which is an integer solution indeed.
If $a=5.$ Then $$4=\left(1+\frac{3}{5}\right)\left(1+\frac{3}{b}\right)\left(1+\frac{3}{c}\right)\leq \left(1+\frac{3}{5}\right)\left(1+\frac{3}{b}\right)^2.$$ We have $$5=a\leq b \leq 5,$$which implies that the possible value of $b$ is $$b=5.$$ If $a=5,b=5$, then $c=\dfrac{16}{3}$, which is not an integer.
As a result, the integer solution is $$a=4,b=5,c=7$$ only. Hence $$a+b+c=4+5+7=16.$$
