Why A.M.$\geq $ G.M. not works here. Find the minimum value of $f(x)= x^2+4x+(4/x)+(1/x^2)$ for $x>0$ 
The minimum value of given function $f(x)= x^2+4x+(4/x)+(1/x^2)$ where $x>0$
(A) 9.5
(B) 10
(C) 15
(D) 20

My try
By A.M G.M inequality
$\frac {x^2+4x+(4/x)+(1/x^2)}{4} \geq \left(x^2.4x. \frac {4}{x}. \frac{1}{x^2}\right)^\frac{1}{4} $
thus minimum value of $f(x)=8$
But by calculus approach the answer is 10.
I am unable to find my mistake where am i doing wrong in A.M. G.M. inequality

AM-GM works but a bit differently:
$$f(x)= x^2+4x+(4/x)+(1/x^2) = x^2+\frac{1}{x^2} + 4(x+\frac{1}{x}) \geq 2 + 4\cdot 2 = 10$$
