Dễ dàng CM đc: $tanA.tanB.tanC=tanA+tanB+tanC$theo BĐT Cauchy có: $P=tan^6A+tan^6B+tan^6C \geq 3.\sqrt[3]{(tanA.tanB.tanC)^6}=3. (tanA.tanB.tanC)^2$
Mà: $tanA+tanB+tanC \geq 3.\sqrt[3]{tanA.tanB.tanC}$
=> $(tanA+tanB+tanC)^3 \geq 27.tanA.tanB.tanC$
=> $(tanA+tanB+tanC)^2 \geq 27 $(do $tanA+tanB+tanC=tanA.tanB.tanC$)
=> $tanA+tanB+tanC \geq 3\sqrt{3}$
=> $P \geq 3.(3\sqrt{3})^2=81$ $(đpcm)$