<s>[INST] 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? 
 [/INST] 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 $\;[3, \infty).$
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