Limitations of Arithmetic-Geometric Inequality? I have to find the range of function $f(x) = x+(1/x) +1$, where $x$ is positive. Now I did it with two ways which we can see below, in equations $(1)$ and $(2)$, by using the AM-GM inequality. 
$$\frac 1 3 \left( x + \frac 1 x + 1 \right) \ge \sqrt[3]{x \cdot \frac 1 x \cdot 1} \implies x + \frac 1 x + 1 \ge 3 \tag 1$$
$$\frac 1 4 \left( x + \frac 1 x + \frac 1 2 + \frac 1 2 \right) \ge \sqrt[4]{x \cdot \frac 1 x \cdot \frac 1 2 \cdot \frac 1 2} \implies x + \frac 1 x + 1 \ge 4 \cdot (1/4)^4 \tag 2$$
So now after seeing these my question is why is my answer different in each case? Are there any limitations of this inequality? 
 A: Your both answers are correct. In first case you have $f(x)\ge 3$ and in second case if you compute you get $4\times (1/4)^{1/4} = 2.828$ which is also true, i.e $f(x)\ge 2.828$.
In first case you have a tighter bound than second case. 
A: The inequation gives you a lower bound, and it has never been said that it was tight. On the opposite, it is known that equality only holds when all arguments are equal.

Notice that


*

*$x=1\implies x=\frac1x=1$ so that $3$ is tight.

*$x=\frac1x=\frac12$ is not possible and $2\sqrt2$ is not tight.
A: look, this isn't a "limitation" of the AM-GM inequality. for example:
$$\frac{1+2}{2} \leq \sqrt{2}$$
and
$$\frac{1+1+1}{3} \leq \sqrt[3]{1}$$
as you can see in the first example the geometric mean isn't certain to be the biggest the amount could be, in your example, the minimum is the first one, but the second one still holds, given it's smaller.
$$(x + \frac{1}{x}+1)'=0$$
$$1-\frac{1}{x^2}=0$$
$$1=\frac{1}{x^2}$$
$$x=1, x=-1 \text{(x is always positive)}$$
$$1+\frac{1}{1}+1=3$$
thus, the minimum is 3.
AM-GM is an inequality, but is in not certain that it is the minimal(or maximal) amount.
A: It is clear that $f(x)$ is unlimited from above. For a lover bound, the following is useful:

For $x>0\;$ is $x+\frac{1}{x}\geq 2,$ with equality at $x=1.$

Therefore, the range of the function $f(x)=x+\frac{1}{x}+1\;$ is $\;[3, \infty).$
