How to show:
In a acute angled $\triangle \ ABC$ show that $\tan(A) \cdot \tan(B)\cdot \tan(C) \geq 3\sqrt{3}$
Any ideas?
How to show:
In a acute angled $\triangle \ ABC$ show that $\tan(A) \cdot \tan(B)\cdot \tan(C) \geq 3\sqrt{3}$
Any ideas?
These may be useful:
Since $\triangle ABC$ is acute, we have $\tan(A),\tan(B),\tan(C)$ positive.
By A.M-G.M you have $\displaystyle \frac{\tan(A)+\tan(B)+\tan(C)}{3} \geq \sqrt[3]{\tan(A)\cdot\tan(B)\cdot\tan(C)}$
$\text{The equality holds if the triangle is}$ $\textbf{equilateral.}$
$A+B=\pi-C$
\begin{align*} &\tan (A+B)= \tan (\pi-C)\\ &(\tan A+ \tan B)/(1-\tan A \tan B)= (\tan \pi- \tan C)/(1+\tan \pi \tan C)=-\tan C\\ &(\tan A+ \tan B)= -\tan C(1-\tan A \tan B)\\ &\tan A + \tan B= -\tan C+ \tan A \tan B \tan C\\ &\tan A + \tan B+ \tan C= \tan A \tan B \tan C \end{align*}
HINT:
Use the AM-GM inequality.