find the minimum value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ If a,b and c are positive real numbers then find the minimum value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ 
Logically,a=b=c=1 but mathematically, I have tried to use some AM-GM, GM-HM inequalities but I am unable to solve this.
 A: This is easy using the AM-GM inequality. We get
$$
\frac{\frac ab + \frac bc + \frac ca}{3}\geq \sqrt[3]{\frac ab\cdot \frac bc \cdot \frac ca} = 1
$$
A: Another approach is to use rearrangement inequality.
$\dfrac ab+\dfrac bc+\dfrac ca\geq\dfrac aa+\dfrac bb+\dfrac cc=3$
A: 
I thought it might be instructive to present a way forward that forgoes appeal to well-know inequalities, but rather relies on straightforward use of calculus.  To that end, we proceed.

Let $x=\frac ab$ and $y=\frac bc$.  Then, we can write
$$\frac ab+\frac bc+\frac ca=x+y+\frac1{xy}$$
Now let $f(x,y)=x+y+\frac1{xy}$.  We see that $$\begin{align}f_1(x,y)&=1-\frac1{yx^2}=0\implies x^2=y \tag 1\\\\
f_2(x,y)&=1-\frac1{xy^2}=0\implies y^2=x \tag 2
\end{align}$$
Putting $(1)$ and $(2)$ together reveals if $f_1(x,y)=f_2(x,y)=0$, then $x=y=1$.  
Now, since $f_{11}(x,y)f_{22}(x,y)-f_{12}^2(x,y)=\frac{3}{(xy)^4}>0$, $f$ attains a local minimum value when $x=y=1$.  Restricting $x>0$ and $y>0$, we see that for positive $x$ and $y$, the local minimum is a global one.
Finally, when $x=y=1$, $a=b=c=1$ and we have

$$\min_{(a,b,c)}\left(\frac ab+\frac bc+\frac ca\right)=1$$ 

for $a>0$, $b>0$, and $c>0$.
A: AM-GM inequality works : let's call $x_1=\frac{a}{b}$, $x_1=\frac{b}{c}$ and $x_3=\frac{c}{a}$. We have
$$\frac{x_1+x_2+x_3}{3}\ge\sqrt[3]{x_1x_2x_3}=1$$
Therefore the minimal value is greater than $3$, and $3$ can be obtained if (and only if) $a=b=c$.
A: Without loss of generality, assume that: $c =min\{a\,\,b\,\,c\}$
$\frac{a}{b} +\frac{b}{c} +\frac{c}{a}$
=$\frac{(a-b)^2}{ab} +\frac{(a-c)(b-c)}{ac} +3 \geq 3$
Or another way! (I will post it soon, now I'm busy)
A: Another way!
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{c^2 (a-b)^2+ab(b-c)^2 +bc(a-c)^2}{abc(b+c)} +3 \geq 3$
Equality holds when $a=b=c$
P/s: I found this identity a long time ago, I don't remember how I found it. Can you help me?
