jmc

First, consider the function $$g(x)=x+\frac{1}{x}.$$If $1\le x<y,$ then $$\begin{array}{rl}g(y)-g(x)& =y+\frac{1}{y}-x-\frac{1}{x}\\ & =y-x+\frac{1}{y}-\frac{1}{x}\\ & =y-x+\frac{x-y}{xy}\\ & =(y-x)\left(1-\frac{1}{xy}\right)\\ & =\frac{(y-x)(xy-1)}{xy}\\ & >0.\end{array}$$Thus, $g(x)$ is increasing on the interval $[1,\infty ).$

By AM-GM (and what we just proved above), $$x+\frac{1}{x}\ge 2,$$so $$g\left(x+\frac{1}{x}\right)\ge 2+\frac{1}{2}=\frac{5}{2}.$$Equality occurs when $x=1,$ to the minimum value of $f(x)$ for $x>0$ is $\boxed{\frac{5}{2}}.$

In particular, we cannot use the following argument: By AM-GM, $$x+\frac{1}{x}+\frac{1}{x+\frac{1}{x}}\ge 2\sqrt{\left(x+\frac{1}{x}\right)\cdot \frac{1}{x+\frac{1}{x}}}=2.$$However, we cannot conclude that the minimum is 2, because equality can occur only when $x+\frac{1}{x}=1,$ and this is not possible.