Dunno leave Loopholes

Applying AM-GM we get

$\frac { f\left( x \right) }{ 2 } \ge \sqrt { \frac { \left\{ x \right\} }{ \left\{ x \right\} } }$

$f\left( x \right) \ge 2\sqrt { \frac { \left\{ x \right\} }{ \left\{ x \right\} } }$

Since $\frac { \left\{ x \right\} }{ \left\{ x \right\} }$ is undefined for $x\in \mathbb Z$ , the minimum value of $f(x)$ can't be determined for $x\in \mathbb R$ .

{x} belongs to [0,1).f(x) is undefined when {x}=0.So we are essentially asked to find min of x+1/x when 0<x<1 . Now f'(x)=1-1/x^2 . It is easy to see this is negative for 0<x<1. So this is a monotonically decreasing function in this open interval.So it will take the minimum value when x=the greatest number in this interval which does not exist(as this is an open interval of R )

We just have to use common sense instead of A.M-G.M inequality because fractional part can be 0.1,0.0001,0.00000000000.........1,

Hence the value of given expression can not be determined

Well this is how I did it .

Let $\{x\} = \frac {p}{q}$ where $gcd(p,q) = 1$ and $q > p$ hence follows that :

$\quad$

$\frac {1} { \{x\} } = \frac {q}{p}\rightarrow \{x\} + \frac {1} { \{x\} } =\frac {p}{q}+\frac {q}{p} = \frac {p^2+q^2}{pq}$

$\quad$

Now,by false assumption, let's suppose there exists a lower limit for the expression.That limit has to be $2$ why? Because $\frac {p^2+q^2}{pq} \ge 2$

$\quad$

But if the limit exists then we must have :

$(p-q)^2 = 0 \rightarrow p = q$ Which contradicts $q > p$ and hence the lower limit cannot be reached.

nice trap , for person who did not check equality condition after using AM-GM inequalty...