Q : Show that in an acute angled triangle ABC, ∑tanA tanB ≥ 9
Sol. We have to prove that ∑tanA tanB ≥ 9
tan A tan B + tan B tan C + tan C tan A ≥ 9
(tan A tan B – 1) + (tan B tan C – 1) + (tan C tan A – 1) ≥ 6
$\displaystyle \frac{tanA + tanB}{tanC} + \frac{tanB + tanC}{tanA} + \frac{tanC + tana}{tanB} \ge 6 $
$\displaystyle (\frac{tanA}{tanC} + \frac{tanC}{tanA}) + (\frac{tanA}{tanB} + \frac{tanB}{tanA}) + (\frac{tanB}{tanC}+ \frac{tanC}{tanB}) \ge 6$
This is true.