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
 A: AM-GM works but a bit differently:
$$f(x)= x^2+4x+(4/x)+(1/x^2) = x^2+\frac{1}{x^2} + 4(x+\frac{1}{x}) \geq 2 + 4\cdot 2 = 10$$
A: The AM-GM inequality does not say that the minimum value is 8. It says that the value is at least 8, hence you know that $\min f(x) \geq 8$. This is consistent with $\min_{x>0} f(x) = 10$. 
You could only say that the minimum value was 8 if the inequality was an equality. In particular, the AM-GM inequality has equality if and only if all the terms are the same. But there is never a point with $x>0$ where they all are equal, i.e. where
$$x^2=4x = 4/x = 1/x^2 $$
A: What you've established is a lower bound for $f(x)$. You have not shown that that is the minimum value that $f(x)$ takes over the domain.
In this case, $f(x)$ takes a minimum value of $10$, which is consistent with the lower bound of $8$, but in actuality, $f(x)$ (for $x>0$) does not ever hit values strictly lower than $10$.
A: The other answers already comment the AM GM inequality, so I shall not comment on it. Instead, I will take a more straightforward, differential calculus approach.
The function can be rewritten as (if we assume $x> 0$)
$$
f(x) = \frac{x^4 + 4x^3  + 4x +1}{x^2}
$$
In this form it's easier to differentiate. The derivative of this function is
$$
f'(x) = \frac{2(x-1)(x+1)^3}{x^3}
$$
Since the function is continuous, the minimum value should be found where $f'(x)=0$. In this case, we only have to check $x = \pm 1$, and we quickly see that $f(1) = 10$ is a minimum value for the function in the defined region $x>0$.
