Mongolian Mathematical Olympiad

Since the inequality is cyclic, we may assume $c=\max (a,b,c)$. Furthermore $a,b,c\in [\frac{1}{3};3]$ implies $\frac{c}{9}\le a\le c$. Substituting $x=\frac{7}{5}-\frac{b}{b+c}$ we get $(1-x)a^2+((2-x)c-xb)a+(1-x)bc\ge 0$ and this is a quadratic inequality with variable $a$.

Graph of this quadratic function is parabola. If $1-x\le 0$ i.e $c\ge \frac{3b}{2}$ then parabola is downward and it is sufficient to check for the values $a=\frac{c}{9},a=c$.

For $a=\frac{c}{9}$ we get $$\begin{array}{rl}\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}& =\frac{c}{9b+c}+\frac{b}{b+c}+\frac{9}{10}=\\ & =\frac{7}{5}+\frac{(3b-c{)}^2}{2(9b+c)(b+c)}\ge \frac{7}{5}\end{array}$$

For $a=c$ we get $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{c}{b+c}+\frac{b}{b+c}+\frac{1}{2}=\frac{3}{2}$$ and we have done.

If $1-x\ge 0$ i.e. $c\le \frac{3b}{2}$ then we get $(2-x)c-xb\ge (2-x)c-xc=2(1-x)c\ge 0$.

Therefore we must prove $$(1-x)a^2+((2-x)c-xb)a+(1-x)bc\ge (1-x)(\frac{c}{9}{)}^2+((2-x)c-xb)\frac{c}{9}+(1-x)bc\ge 0$$ This is same as we done above in the case $a=\frac{c}{9}$. This completes the proof. Equality holds iff $(1;\frac{1}{3};3)$, $(\frac{1}{3};3;1)$, $(1;3;\frac{1}{3})$.