$a,b,c \in \mathbb{R}^+, \dfrac{a}{a+7b}+\dfrac{b}{b+7c}+\dfrac{c}{c+7a} \geq \dfrac{3}{8}$ How to prove
$a,b,c \in \mathbb{R}^+$
$$\dfrac{a}{a+7b}+\dfrac{b}{b+7c}+\dfrac{c}{c+7a} \geq \dfrac{3}{8}$$
my solution 
this equivalent to
$$\dfrac{7(13a^2b+13b^2c+13c^2a+35a^2b^2+35b^2c^2+35b^2c^2-144abc)}{8(a+7b)(b+7c)(c+7a)} \geq 0$$
from AM-GM, we can get 
$$13a^2b+13b^2c+13c^2a \geq 39abc$$
$$35b^2c^2+35b^2c^2+35b^2c^2 \geq 105abc$$
done
I want to know better way to prove this

Your solution is correct.
Let $A$ be your expression and $B=a(a+7b) + b(b+7c) +c(c+7a)$. Then it is enought, by Cauchy Shwarz inequality, to prove $$8(a+b+c)^2\geq 3(a^2+b^2+c^2+7ab+7bc+7ca)$$
i.e.
$$5(a^2+b^2+c^2)\geq 5(ab+bc+ca)$$ which is true by again CS or AG.
