Prove $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ Prove that $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ with a,b,c > 0 

Hint:
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-3=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{b}{a}-1\right)=\dfrac{(a-b)^2}{ab}+\dfrac{(a-c)(b-c)}{ac}\tag{1}$$

 Similarly, $$\dfrac{a+c}{b+c}+\dfrac{b+c}{a+b}+\dfrac{a+b}{a+c}-3=\dfrac{(a-b)^2}{(b+c)(a+b)}+\dfrac{(a-c)(b-c)}{(b+c)(a+c)}\tag{2}$$
 As $a,b,c >0$, it is obvious that $(1)$ is greater than $(2)$ as without loss of generality, we can set $c=\max{\{a,b,c\}}$ or $c=\min{\{a,b,c\}}$

