Let $a,b,c>0$ then prove this inequality holds $$\left( \frac{a}{b+c}+\frac{b}{c+a} \right)\left( \frac{b}{c+a}+\frac{c}{a+b} \right)\left( \frac{c}{a+b}+\frac{a}{b+c} \right)\ge 1+\frac{1}{8}{{\left( \frac{a-b}{a+b} \right)}^{2}}{{\left( \frac{b-c}{b+c} \right)}^{2}}{{\left( \frac{c-a}{c+a} \right)}^{2}}$$
I broke the whole expression then expressed it like this
\begin{align*}
LHS-RHS & =\frac{\sum \limits_{cyc}{c(10a^2b+10ab^2+11abc+c^3)(a-b)^2}+\sum \limits_{cyc}{ab[(a^2-6ab+ac+2b^2+5bc-3c^2)^2+(2a^2-6ab+5ac+b^2+bc-3c^2)^2]}}{5(b+c)^2(c+a)^2(a+b)^2} \ge {0}
\end{align*}
But this proof is very ugly. Do you have any better easy and beautiful proof
 A: Let $\frac{a}{b+c}=\frac{x}{2},$ $\frac{b}{a+c}=\frac{y}{2}$ and $\frac{c}{a+b}=\frac{z}{2}.$
Thus, since $$\frac{x-y}{x+2}=\frac{\frac{2a}{b+c}-\frac{2b}{a+c}}{\frac{2a}{b+c}+2}=\frac{a-b}{a+c},$$ we need to prove that:
$$\prod_{cyc}(x+y)\geq8+\frac{\prod\limits_{cyc}(x-y)^2}{\prod\limits_{cyc}(x+2)^2}.$$
Also, $$2=\sum_{cyc}\frac{b+c}{a+b+c}=\sum_{cyc}\frac{1}{1+\frac{a}{b+c}}=\sum_{cyc}\frac{1}{1+\frac{x}{2}}=2\sum_{cyc}\frac{1}{x+2},$$ which gives
$$xy+xz+yz+xyz=4.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $u$ and we need to prove 
$$9uv^2-w^3\geq8+\frac{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}{(w^3+6v^2+12u+8)^2}$$ or $f(u)\geq0,$ where
$$f(u)=(9uv^2-w^3-8)(w^3+6v^2+12u+8)^2-27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6).$$
But $$f'(u)=9v^2(w^3+6v^2+12u+8)^2+$$
$$+24(9uv^2-w^3-8)(w^3+6v^2+12u+8)-27(6uv^4-12u^2w^3+6v^2w^3)\geq$$
$$\geq9(v^2(w^3+6v^2+12u+8)^2-18uv^4)\geq0,$$ which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $u$, which by $uvw$ happens for equality case of two variables. 
Let $x=y$. 
Thus, $a=b$ and since in this case our inequality is obvious, we are done!
