# 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 inequality says that $f(x) \geq 8$. But it does not follow that the minimum value is 8. Apr 26, 2018 at 7:40

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

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 $$x^2=4x = 4/x = 1/x^2$$

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 $10$.

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 $x>0$.