Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle. The question is:

If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$

My solution: 
$(\cot A + \cot B + \cot C)^2\ge0$   //  square of a real number
$\implies \cot^2A +\cot^2B + \cot^2 +2 \ge0   $  //Conditional identity used: $\cot A \cot B + \cot B \cot C + \cot A \cot C =1$
$\implies \cot^2A +\cot^2B + \cot^2 C \ge -2$
Thus according to me the answer should be $-2$. However, the answer key states that the answer is $1$. Where have I gone wrong? 

it is equivalent to
$$\frac{1}{\sin(A)^2}+\frac{1}{\sin(B)^2}+\frac{1}{\sin(C)^2}\geq 4$$
with $$\sin(A)=\frac{a}{2R}$$ etc and $$S=\sqrt{s(s-a)(s-b)(s-c)}$$ and $$S=\frac{abc}{4R}$$ we get
$$b^2c^2+c^2a^2+a^2b^2-(-a+b+c)(a-b+c)(a+b-c)(a+b+c)\geq 0$$
and this is equivalent to
$$a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2$$
which is true.
