Trigonometry and triangle proof Question:
Prove that in an acute angle triangle ABC:
$$\tan A\tan B +\tan A \tan C + \tan B \tan C \geq 9$$
I have no idea where to even begin this question. Please help me!

Since for an acute triangle $ABC$ we have
$$\begin{align}\tan A+\tan B&=(\tan (A+B))\cdot (1-\tan A\tan B)\\&=(\tan(\pi -C))\cdot (1-\tan A\tan B)\\&=(-\tan C)\cdot (1-\tan A\tan B),\end{align}$$
we have$$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$
Hence, we have, by AM-GM inequality,$$\tan A+\tan B+\tan C\ge 3\sqrt[3]{\tan A\tan B\tan C}$$
$$\iff \tan A\tan B\tan C\ge 3\sqrt[3]{\tan A\tan B\tan C}$$
$$\iff \tan A\tan B\tan C\ge 3\sqrt 3.$$
Hence, we have, by AM-GM inequality,
$$\begin{align}\tan A\tan B+\tan B\tan C+\tan C\tan A&\ge 3\sqrt[3]{\tan^2A\tan^2B\tan^2C}\ge 9.\end{align}$$
