How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
? 
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue from here. 
Also I try Cauchy-Schwarz: $$\sum_{cyc} \frac{a}{7a+b}\le\sqrt{a^2+b^2+c^2}\sqrt{\sum_{cyc} \frac{1}{(7a+b)^2}}.$$
Now what?
 A: By C-S
$$\sum_{cyc}\frac{a}{7a+b}=\frac{3}{7}+\sum_{cyc}\left(\frac{a}{7a+b}-\frac{1}{7}\right)=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b}{7a+b}=$$
$$=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$
Id est, it's enough to prove that
$$\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}\leq\frac{3}{8}$$ or
$$8(a+b+c)^2\geq3\sum\limits_{cyc}(7ab+a^2)$$ or $$\sum_{cyc}(a-b)^2\geq0$$ and we are done!
A: By AM-GM we have $$a^2b+ac^2+b^2c\geq3abc$$ and $$a^2c+ab^2+bc^2\geq 3abc$$ so that $$35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)\geq 3(35+13)abc=144abc.$$
Now, note that $$\frac38-\sum_{\text{cyc}} \frac{a}{7a+b}=\frac{35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)-144abc}{8 (7 a+b) (a+7
   c) (7 b+c)},$$
which is non-negative by the previous result.
We have equality if and only if we have equality in both AM-GMs which implies $a=b=c$.
A: Can be much detail, please
$$\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$
