## Trigonometric Optimization continued

Prove that in any acute angled triangle ABC, $\tan {A}+\tan{B}+\tan{C} \geq 3\sqrt{3}$ with equality holding if and only if the triangle is equilateral. (IITJEE 1998)

Proof:

Suggestion: Try this without reading further! It looks complicated, but need not be so!! Then, after you have attempted whole-heartedly, compare your solution with the one below.

The solution to the above problem is based on the well-known identity:

$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$. For brevity, denote $\tan{A}, \tan{B}, \tan{C}$

by x, y and z respectively. As ABC is acute-angled, x, y, z are all positive and the AM-GM inequality which says

$x+y+z \geq 3{(xyz)}^{1/3}$ can be applied. Taking cubes of both the sides and cancelling

$x+y+z$, (which is positive) this gives $(x+y+z)^{2} \geq 27$. Taking square root we get the desired inequality. If equality is to hold, then it must also hold in the AM-GM inequality, which can happen if and only if

$x=y=z$, that is, if and only if the triangle is equilateral.

Still, this approach requires some caution. Actually, there are so many trigonometric identities that there is no unanimity as to which ones among them are standard enough to be assumed without proof !! But, of course, the IITJEE Examinations, both Mains and Advanced are multiple choice only.

More later,

Nalin Pithwa

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