Question

If A, B, C are the angles of an acute angled triangle, then show that tan A . tan B . tan C ≥ 3√3.

Answer 1

A, B, C being the angles of an acute angled triangle, tan A > 0, tan B > 0, tan C > 0. 

Also,  A + B + C = π

⇒ A + B = π – C ⇒ tan (A + B) = tan (π – C)

⇒ \(\frac{tan\,A+tan\,B}{1-tan\,Atan\,B}\) = - tan C ⇒ tan A + tan B + tan C = tan A . tan B . tan C

Now applying AM > GM, we have

\(\frac{tan\,A+tan\,B+tan\,C}{3}\) > (tan A . tan B . tan C)^\(\frac13\) 

⇒ (tan A + tan B + tan C)³ > 27 (tan A . tan B . tan C) 

⇒ (tan A . tan B . tan C)³ > 27 (tan A . tan B . tan C)       (From (i)) 

⇒ (tan A . tan B . tan C)² > 27 ⇒ tan A . tan B . tan C > 3√3 .