Trigonometry and triangle proof

Question

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!

Answer 1

Note that $$\tan C=-\tan(A+B)=\frac{\tan A+\tan B}{\tan A\tan B-1}.$$ That means $$\tan A+\tan B+\tan C=\tan A\tan B\tan C,$$ or $$\sum_{cyc}\frac{1}{\tan A\tan B}=1.$$ The given inequality is now obvious.

Answer 2

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, $$\begin{align}\tan A\tan B+\tan B\tan C+\tan C\tan A&\ge 3\sqrt[3]{\tan^2A\tan^2B\tan^2C}\ge 9.\end{align}$$