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 $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$?
 A: Assume $$\dfrac{a}{b}=x,\dfrac{b}{c}=y,\dfrac{c}{a}=z$$
So for instance
$$\dfrac{a+c}{b+c}=\dfrac{1+xy}{1+x}=x+\dfrac{1-x}{1+y}$$
And the problem would be transformed to:
$$\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1}\ge0$$
$\equiv(x^2-1)(z+1)+(y^2-1)(x+1)+(z^2-1)(y+1)\ge0$
$\equiv \sum{x^2z}+\sum{x^2}\ge\sum{x}+3$
We have $xyz=1$, hence :$$\sum{x^2z}\ge3$$
And also $x+y+z\ge3$, so $$\sum{x^2}\ge\dfrac{(\sum{x})^2}{3}\ge\sum{x}$$
Problem solved
A: our inequality is equivalent to $$a^4c^2+a^2b^4+b^2c^4+a^3b^3+a^3c^3+b^3c^3\geq a^3bc^2+a^2b^3c+ab^2c^3+3a^2b^2c^2 $$we have $$a^3b^3+a^3c^3+b^3c^3\geq 3a^2b^2c^2$$ by AM-GM, and  the rest is equivalent to
$$(a-b)c^2(a^3-b^2c)+(b-c)b^2(a^2b-c^3)\geq 0$$ if we assume that $$a\geq b\geq c$$
and if $$a\geq c\geq b$$ on obtain
$$(a-c)c^2(a^3-b^2c)+(c-b)a^2(ac^2-b^3)\geq 0$$
thank you Michael Rozenberg!
A: 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. 
A: Note that, $0=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}=\ln\dfrac{a+b}{a+c}+\ln\dfrac{b+c}{b+a}+\ln\dfrac{c+a}{c+b}$ and taking $\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$ WLOG gives us $ac\ge b^2$ ,$\;$ $a^2\ge bc$ $\;$ and $\;$ $ab\ge c^2$ also because $(a+c)^2=a^2+c^2+ac+ac\ge \dfrac{(a+c)^2}{2}+b^2+ac\ge(a+c)b+b^2+ac=(a+b)(b+c)$
$$\ln\dfrac{c+a}{c+b}\ge\ln\dfrac{a+b}{a+c}\ge\ln\dfrac{b+c}{b+a}$$ $\;$ or 
$$\ln\dfrac{c+a}{c+b}\ge\ln\dfrac{b+c}{b+a}\ge\ln\dfrac{a+b}{a+c}$$ must be satisfied.
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Also it is easy to show $\dfrac{a}{b}\ge\dfrac{c+a}{c+b}$ because $ab+ac\ge ab+bc$ and $\dfrac{c}{a}\le\dfrac{b+c}{b+a}$ because $bc+ca\le ab+ac$ also $\dfrac{c}{a}\le\dfrac{a+b}{a+c}$ because $a^2\ge ac$ and $ab\ge c^2$. 
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So; $$\left(\ln\dfrac{c+a}{c+b}, \ln\dfrac{a+b}{a+c}, \ln\dfrac{b+c}{b+a}\right) \prec  \left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)$$
or $$\left(\ln\dfrac{c+a}{c+b}, \ln\dfrac{b+c}{b+a}, \ln\dfrac{a+b}{a+c}\right) \prec  \left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)$$
 and by Karamata's inequality with using 
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$f(x)=e^x$ ($f''(x)=e^x>0$) $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$$
