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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?

  • 1
    The question http://math.stackexchange.com/questions/8732/how-to-prove-this-trignometrical-identities includes the identity Srivatsan mentions.2011-08-13

3 Answers 3

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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 $\displaystyle \frac{\tan(A)+\tan(B)+\tan(C)}{3} \geq \sqrt[3]{\tan(A)\cdot\tan(B)\cdot\tan(C)}$

  • $\text{The equality holds if the triangle is}$ $\textbf{equilateral.}$

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$A+B=\pi-C$

\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*}

  • 1
    $A+B+C=\pi$, not $2\pi$. Fortunately, $\tan(\pi)=\tan(2\pi)$.2011-08-13
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HINT:

Use the AM-GM inequality.