Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$ 
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$

What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} = \dfrac{a^2c+b^2a+c^2b}{abc} \geq \dfrac{3abc}{abc} = 3$. Then how can I use this to prove that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$?

Without Loss of Generality, Let us assume that $c$ is the maximum of $a,b,c$
Notice that $$\sum _{ cyc }^{  }{ \frac { a }{ b }  } -3=\frac { a }{ b } +\frac { b }{ a } -2+\frac { b }{ c } +\frac { c }{ a } +\frac { b }{ a } -1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}$$
However, since $(a-b)^2,(c-a)(c-b)\ge 0$, $$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac} \ge \frac{(a-b)^2}{(c+a)(c+b)}+\frac{(c-a)(c-b)}{(a+c)(b+a)}=\sum _{ cyc }^{  }{ \frac { a+b }{ c+a }  }-3$$ 
Our proof is done. 
