Question

If $a,b,c,d\in R^{+}$ such that $a+b+c+d=4\sqrt{3}$ and $abcd+9=ab+ac+ad+bc+bd+cd$, then the number of solutions is

Answer 1

$\text{Using A.M. - G.M. inequality,}\frac{ab+ac+ad+bc+bd+cd}{6}\ge \sqrt[6]{6ab\times ac\times ad\times bc\times bd\times cd}\Rightarrow ab+ac+ad+bc+bd+cd\ge 6\sqrt{abcd}\text{Now},abcd+9=ab+ac+ad+bc+bd+cd\Rightarrow abcd+9\ge 6\sqrt{abcd}\Rightarrow abcd-6\sqrt{abcd}+9\ge 0\Rightarrow (\sqrt{abcd}-3)\ge 0\Rightarrow \sqrt{abcd}\ge 3...\left(1\right)\text{Again},\frac{a+b+c+d}{4}\ge \sqrt[4]{4abcd}\Rightarrow \frac{4\sqrt{3}}{4}\ge \sqrt[4]{4abcd}\Rightarrow 3\ge \sqrt{abcd}...\left(2\right)\text{From equation (1) and (2), we have}\sqrt{abcd}=3\text{And this equality only holds when}a=b=c=d=\sqrt{3}\therefore \text{There is only one solution i.e.}(\sqrt{3},\sqrt{3},\sqrt{3},\sqrt{3}).$