Question

If $△ABC$ is an acute angle triangle, then the minimum value of $\tan A+\tan B+\tan C$ is

• A. $2\sqrt{2}$
• B. $3\sqrt{2}$
• C. $3\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

The correct option is C $3\sqrt{3}$

In a $△ABC$, we know that
$\tan A+\tan B+\tan C=\tan A\tan B\tan C\cdots \left(1\right)$

As $\tan A,\tan B,\tan C>0$, so using
$A.M.\ge G.M.$, we get
$\tan A+\tan B+\tan C\ge 3(\tan A\tan B\tan C{)}^{1/3}$
From equation $\left(1\right)$,
$\tan A+\tan B+\tan C\ge 3(\tan A+\tan B+\tan C{)}^{1/3}\Rightarrow (\tan A+\tan B+\tan C{)}^{2/3}\ge 3\Rightarrow \tan A+\tan B+\tan C\ge 3^{3/2}\therefore \tan A+\tan B+\tan C\ge 3\sqrt{3}$

Hence, the minimum value of the given expression is $3\sqrt{3}$.