Question

Proove that in ∆ABC tanA + tanB + tanC
$\geqslant 3 \sqrt{3}$
where A, B ,C are acute angles​

Answer 1

Hi !!

Solution:- In ∆ABC ,

tanA - tanB + tanC = tanA*tanB*tanC

Also, tanA + tanB + tanC/3

$\geqslant 3 \sqrt{tanA + tanB + tanC}$

[Since, A.M greater than or equal to G.M ]

=> tanA tanB tanC

$\geqslant \: 3 \times \sqrt[3]{tanA\: tanB \: tanC}$

=> tan²A tan²B tan²C

$\geqslant 27$

[Cubing both side ]

=> tanA tanB tanC

$\geqslant \sqrt[3]{3}$

=> tanA + tanB+ tanC

$\geqslant \ \sqrt[3]{3}$

____________________________

Hope it helps you !!

@Raj❤

Answer 2

hope it helps!

mark as brainliest!