Question

In an acute angled triangle ABC, the minimum value of $ta{n}^{n}A+ta{n}^{n}B+ta{n}^{n}C$. is
(When $nϵN,n>1)$

• A. $3^{\frac{n}{2}}$
• B. $3^n$
• C. $3^{\frac{n}{2}+1}$
• D. $3^{\frac{n}{2}-1}$

Answer 1

The correct option is C $3^{\frac{n}{2}+1}$

$\frac{ta{n}^{n}A+ta{n}^{n}B+ta{n}^{n}C}{3}\ge (\frac{tanA+tanB+tanC}{3}{)}^n$
(Arithmetic mean of $n^{th}$ power of numbers)

$\frac{ta{n}^n+ta{n}^{n}B+ta{n}^{n}C}{3}\ge (\frac{tanA+tanB+tanC}{3}{)}^n$
(Since $tanA+tanB+tanC\ge 3\sqrt{3})\Rightarrow ta{n}^{n}A+ta{n}^{n}B+ta{n}^{n}C\ge 3^{\frac{n}{2}+1}$

Hence, option 'C' is correct.