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? 

Using Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,
$$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$
Now $$(\cot A-\cot B)^2+(\cot B-\cot C)^2+(\cot C-\cot A)^2\ge0$$
$$\iff\cot^2A+\cot^2B+\cot^2C\ge\cot A\cot B+\cot B\cot C+\cot C\cot A$$
