Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$

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

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!