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
On
Better way is always calculus but you may need Wolfram Alpha if you are not into solving equations.
Let $\frac{b}{a}=x$ and $\frac{c}{b}=y$. We want to maximize the function $f(x,y)=\frac{1}{1+7x}+\frac{1}{1+7y}+\frac{xy}{xy+7}$ in the open first quadrant of the cartesian plane.
Solving $f_x=0$ and $f_y=0$, I think, we get $(x,y)=(49,\frac{1}{2041}), (\frac{1}{2041},49)$ or $(1,1)$.
$(1,1)$ will do the minimum.
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.