Question

If A, B, and C are acute positive angles such that $A + B + C = \pi$ and $\cot A\cot B\cot C = K$, then
$1)K \leqslant \dfrac{1}{{3\sqrt 3 }}$
$2)K \geqslant \dfrac{1}{{3\sqrt 3 }}$
$3)K < \dfrac{1}{9}$
$4)K > \dfrac{1}{3}$

Answer 1

Hint : We have been given three acute positive angles. The conditions given for these angles are $A + B + C = \pi$ and $\cot A\cot B\cot C = K$. We shall use the $\tan (A + B + C)$ formula so that we can substitute the given data in it. Then we apply the inequality for Arithmetic Mean and Geometric Mean. Since the relations are in$\tan$we convert it to $\cot$to use find out the value of
The formulas used to solve the problem are:
$\tan \theta = \dfrac{1}{{\cot \theta }}$
$\tan \left( {A + B + C} \right) = \dfrac{{\left( {\tan A + \tan B + \tan C - \tan A\tan B\tan C} \right)}}{{\left( {1 - \tan A\tan B - \tan B\tan C - \tan C\tan A} \right)}}$
The AM GM inequality which states that Arithmetic Mean is greater than or equal to Geometric Mean for a given list of non-negative real numbers.

Complete step-by-step answer :
We consider the formula,
$\tan \left( {A + B + C} \right) = \dfrac{{\left( {\tan A + \tan B + \tan C - \tan A\tan B\tan C} \right)}}{{\left( {1 - \tan A\tan B - \tan B\tan C - \tan C\tan A} \right)}}$
It is given that,
$A + B + C = \pi$
So, we substitute the above in the formula,
$\Rightarrow \tan (A + B + C) = \tan \pi$
We know that,
$\tan \pi = 0$
Now the equation becomes,
$\Rightarrow 0 = \dfrac{{\left( {\tan A + \tan B + \tan C - \tan A\tan B\tan C} \right)}}{{\left( {1 - \tan A\tan B - \tan B\tan C - \tan C\tan A} \right)}}$
By rearranging, we get
$\Rightarrow \left( {1 - \tan A\tan B - \tan B\tan C - \tan C\tan A} \right)\left( 0 \right) = \left( {\tan A + \tan B + \tan C - \tan A\tan B\tan C} \right)$
We are only left with the LHS, so we rearrange it as,
$\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$
We know that, $\tan \theta$ can be written as $\dfrac{1}{{\cot \theta }}$
Substituting this in the above equation, we get,
$\Rightarrow \tan A + \tan B + \tan C = \dfrac{1}{{\cot A\cot B\cot C}}$……… (1)
It is given that all angles are acute, hence, we can apply the AM-GM inequality for equation (1)
The AM is given by,
$\dfrac{{\tan A + \tan B + \tan C}}{3}$
The GM is given by,
${\left( {\tan A\tan B\tan C} \right)^{\dfrac{1}{3}}}$
The inequality states that $AM \geqslant GM$
So, we get,
$\Rightarrow \dfrac{{\tan A + \tan B + \tan C}}{3} \geqslant {\left( {\tan A\tan B\tan C} \right)^{\dfrac{1}{3}}}$
From (1), it can be written as
$\Rightarrow \dfrac{1}{{3\left( {\cot A\cot B\cot C} \right)}} \geqslant {\left( {\dfrac{1}{{\cot A\cot B\cot C}}} \right)^{\dfrac{1}{3}}}$
It is given that, $\cot A\cot B\cot C = K$
$\Rightarrow \dfrac{1}{{3K}} \geqslant {\left( {\dfrac{1}{K}} \right)^{\dfrac{1}{3}}}$
$\Rightarrow \dfrac{1}{{3K}} \geqslant {\left( K \right)^{\dfrac{{ - 1}}{3}}}$
Now, by simplifying,
$\Rightarrow \dfrac{1}{{3K}} \geqslant \dfrac{1}{{\sqrt {3K} }}$
The inequality changes when the terms are taken from LHS to RHS
$\Rightarrow K \geqslant \dfrac{1}{{3\sqrt 3 }}$
The final answer is $K \geqslant \dfrac{1}{{3\sqrt 3 }}$
Hence, option (2) is the correct answer.

Note : Take the formulas according to the given question. Be careful while taking the cube root. Note that the inequality sign changes as you move terms from LHS to RHS or vice versa. We make use of the given data into an already existing formula, so be careful to use the given data after mentioning the formula.