If A, B, C are acute positive angles such that A+B+C=π and cotA cotB cotC=K, then
We have,
tan(A+B+C)=(tanA+tanB+tanC−tanAtanBtanC)(1−tanAtanB−tanBtanC−tanCtanA)
A+B+C=π
⇒tan(A+B+C)=tanπ=0
⇒tanA+tanB+tanC=tanAtanBtanC
....(1)
As all angles are acute, we can apply the A.M.≥G.M. inequality for tanA,tanB,tanC
tanA+tanB+tanC3≥(tanAtanBtanC)1/3
Using equation (1), and cotAcotBcotC=K
⇒13K≥(K)−1/3
⇒K≤13√3
⇒cotAcotBcotC≤13√3