Find all solutions of $a,b,c,d$ if $a+b+c+d=12$ and $abcd = 27 +ab+ac+ad+bc+bd+cd$ Find all solutions of $a,b,c,d>0$ if
$$a+b+c+d=12$$
$$abcd = 27 +ab+ac+ad+bc+bd+cd$$

Attempt:
$a + b + c + d  = 12 \implies 3 \ge  \sqrt[4]{abcd} \implies 81 \ge abcd$ by AM-GM, eqyality when $a=b=c=d$. Also
$$ abcd - 27 = ab + ac + ad + bc + bd + cd \ge 6 \sqrt[6]{a^{3} b^{3} c^{3} d^{3}} = 6 \sqrt{a b c d} $$
$$ (abcd - 27)^{2} \ge 36 a b c d$$
dari AM-GM, equality when $ab=ac=ad=bc=bd=cd$. then we must have
$$ (abcd - 27)^{2} = 36 \times 81 \implies abcd = 54 + 27 = 81 $$
atau
$$ (abcd - 27)^{2} = 36 \times 81 \implies abcd = -54 + 27 = -27 (\text{not possible}) $$
So $abcd = 81$. Since $a=b=c=d$ then $a=b=c=d=3$.
Is this the only solution? if it is, why?
 A: You are fine : essentially, you got $$a+b+c+d \geq 4 \sqrt[4]{abcd} \implies 81 \geq abcd$$
from the first line. Then from the next few, you got :
$$
ab+bc+cd+da+ac+bd \geq 6 \sqrt[6]{a^3b^3c^3d^3} = 6 \sqrt{abcd} 
$$
Combining this with $ab+bc+cd+da+ac+bd = abcd - 27$ gives that $abcd - 27 \geq 6 \sqrt{abcd}$, which can be rearranged to $(\sqrt{abcd} - 3)^2 \geq 36$. Now from here and the fact that $\sqrt{abcd}$ must be positive we get $\sqrt{abcd} \geq 9$. Combining with our first observation we get that equality is attained in AM-GM which gives $a=b=c=d=3$.
A: I think it's better from $$abcd-27\geq6\sqrt{abcd}$$ to write
$$abcd-6\sqrt{abcd}-27\geq0$$ or
$$(\sqrt{abcd}-9)(\sqrt{abcd}+3)\geq0$$ or
$$abcd\geq81$$ and since by your work
$$abcd\leq81,$$ it's enough to check an equality case.
Another solution.
By AM-GM we obtain:
$$abcd=27\left(\frac{a+b+c+d}{12}\right)^4+(ab+ac+bc+ad+bd+cd)\left(\frac{a+b+c+d}{12}\right)^2\geq$$
$$\geq27\left(\frac{4\sqrt[4]{abcd}}{12}\right)^4+6\sqrt[6]{a^3b^3c^3d^3}\left(\frac{4\sqrt[4]{abcd}}{12}\right)^2=abcd,$$
which says that the inequality is equality, which occurs for $a=b=c=d,$ which gives only
$$(a,b,c,d)=(3,3,3,3).$$
