find the min of

Simply applying AM-GM inequality,we get

$(a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{a}) \ge (2\sqrt{\frac{a}{b}}*2\sqrt{\frac{b}{c}}*2\sqrt{\frac{c}{a}})$

which results in $\ge 8$ .Hence 8 is the answer.

Applying Hölder's inequality and then AM-GM inequality :

$\left(a+\frac 1b\right)\left(b+\frac 1c\right)\left(c+\frac 1a\right) \ge \left(abc + \frac 1{abc} \right)^3 \ge \left(2\sqrt{\frac {abc}{abc}}\right)^3 = 2^3 = \boxed 8$

Equality occurs when $a=b=c=1$ .

$(a+\dfrac{1}{b})+(b+\dfrac{1}{c})+(c+\dfrac{1}{a})=\left (abc+\dfrac{1}{abc}\right )+(a+b+c)+\left (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right )$ .

Now,

$abc+\dfrac{1}{abc}\geq 2$ (A. M.-G. M. inequality) .

$(a+b+c)+\left (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right ) \geq 2\sqrt {(a+b+c)\left (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right ) }$ (A. M.-G. M. inequality) , $\geq 2\sqrt 9=6$ (A. M.-H. M. inequality)

So the minimum of the given expression is $2+6=\boxed 8$ .

Lemma: First, consider the function $f(x) = x + \frac{1}{x}$ . The minimum value of this function, for $x \in \mathbb{R}^{+},$ is found at $x=1$ , where $f(x) = 2$ .

(This is an easy fact to note considering the behavior of $f'(x) = 1 - \frac{1}{x^2}.)$

Now expand the expression above:

$(a + \frac{1}{b})(b + \frac{1}{c})(c + \frac{1}{a}) =$

$abc + a + b + c + \frac{1}{a} + \frac{1}{b} +\frac{1}{c} + \frac{1}{abc} =$

$(a + \frac{1}{a}) + (b + \frac{1}{b}) + (c + \frac{1}{c}) + (abc + \frac{1}{abc} )$

$\geq$ (by Lemma) $2 + 2 + 2 +2 = 8.$

And the minimum is achieved at $a=b=c=1$ .

Expanding, we get the expression $\displaystyle abc + a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc}$ .

We re-arrange this to get the expression $\displaystyle abc + \frac{1}{abc} + a + \frac{1}{a} + b + \frac{1}{b} + c + \frac{1}{c}$ .

Then, we apply AM-GM. In general, for positive reals (which is the domain stated in the question):

$\displaystyle \frac{k + \frac{1}{k}}{2} \ge \sqrt{k \times \frac{1}{k}}$

$\displaystyle k + \frac{1}{k} \ge 2$

We apply this logic to our equation, and the minimum value of it is $\boxed{8}$ . Equality occurs when $a = b = c = 1$ .