A Trigonometric Inequality

Another method, albeit more rigorous, is to use LaGrange Multipliers. Let $f(A,B,C) = cot^2(A) + cot^2(B) + cot^2(C)$ and $g(A,B,C) = A + B + C = \pi$ . Taking $\nabla f = \lambda \cdot \nabla g$ yields:

$\lambda = -2cot(A)csc^2(A);$

$\lambda = -2cot(B)csc^2(B);$

$\lambda = -2cot(C)csc^2(C);$

which equality holds if and only if $A = B = C = \frac{\pi}{3}.$ (i.e. an equilateral triangle). Next, we calculate the $3 \times 3$ symmetric Hessian matrix of $f(A,B,C) \Rightarrow F(A,B,C):$

$F = \begin{bmatrix}{2csc^4(A) + 4cot^2(A)csc^2(A)} && {0} && {0} \\ {0} && {2csc^2(B) + 4cot^2(B)csc^2(B)} && {0} \\ {0} && {0} && {2csc^4(C) + 4cot^2(C)csc^2(C)}\end{bmatrix}$

and which $F(\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{3}) = \frac{48}{9} \cdot I_3$ (where $I_3$ is the $3 \times 3$ identity matrix) is positive-definite $\Rightarrow$ $f(A,B,C)$ is globally minimized at ( $\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{3})$ , or $f(\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{3}) = 3 \cdot cot^2(\frac{\pi}{3}) = 3 \cdot (\frac{1}{\sqrt3})^2 = \boxed{1}.$