$\displaystyle\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right) \ge \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$
,but i don't prove that!, you have one hint?
thanks very much
i solve this$\displaystyle\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right) \ge \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$
with that
$$(a+b+c)^2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-10(a^2+b^2+c^2)+(ab+bc+ca)=\sum \frac{(a-b)^2(b-2c)^2}{bc} $$
but is very ugly (how we find this decomposition?)
thanks very much