Minimum value of trigonometric equation Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it is wrong, but why?

$$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x}$$ 
or $$y=\frac{16}{4}*\sec^2 x \csc^2 x-8 +\frac{8}{4}\frac{\cos^{4} x}{\sin^2 x* \cos^2 x}$$ 
or  
$$y=4(\tan^2 x +1)(\cot^2 x +1)-8+2 \cot^2 x$$
$$y=4(\tan^2 x + \cot^2 x + 2)-8+2 \cot^2 x$$   
Which after simplifying gives,
$$y=4\tan^2 x + 6\cot^2 x$$
Edit: as user suggested the answer in comment is not valid. But one here can easily use AM-GM inequality to reach at correct answer
