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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?

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Use $\tan A \cdot \tan B \cdot \tan C = \tan A + \tan B + \tan C$, and the AM-GM inequality. – Srivatsan Aug 13 '11 at 6:45
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The question math.stackexchange.com/questions/8732/… includes the identity Srivatsan mentions. – Jonas Meyer Aug 13 '11 at 6:56

3 Answers

up vote 1 down vote

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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up vote 0 down vote

HINT:

Use the AM-GM inequality.

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up vote 0 down vote

A+B+C=2PI

 A+B=PI-C
    tan (A+B)= tan (PI-C)
    (tan A+ tan B)/(1-tan A tanB)= (tan PI- tan C)/(1+tan PI tan C)=-tanC
    (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
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Now u can use the am-gm inequality as others have stated. and no problem in using that inequality as all the variables involved iwll be positive quantities in case of an acute angled traingle – Bhargav Aug 13 '11 at 6:52
You mean $A+B=\pi - C$, I guess. It would also be good if you indicated where exactly you need to use that $A,B,C \lt \pi/2$. – t.b. Aug 13 '11 at 6:54
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$A+B+C=\pi$, not $2\pi$. Fortunately, $\tan(\pi)=\tan(2\pi)$. – Jonas Meyer Aug 13 '11 at 6:55

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