Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$ Reducing this whole expression i finally came to this
$$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$
Here I am stuck. I can't prove this. 
So I thought maybe I should try in another way.
Let $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$
Hence the whole expression comes to this inequality
\begin{align*} (a+b+c)^2&(a+b)(b+c)(c+a) \\
& \geq 4(ab+bc+ca)(ab^2+a^2b+bc^2+b^2c+ca^2+c^2a)\\
\implies   (a+b+c)^2&\left((a+b+c)(ab+bc+ca)-abc\right)\\
&  \geq 4(ab+bc+ca)\left((a+b+c)(ab+bc+ca)-3abc\right)\\
\implies  (3u)^2\left(3u\times3v^2-w^3\right)&\geq 4\times3v^2\left(3u\times3v^2-3w^3\right)\\
\implies  9u^2(9uv^2-w^3)&\geq 12v^2(9uv^2-3w^3)\\
\implies  9u^3v^2-u^2w^3&\geq 12uv^4-4v^2w^3\end{align*}
Here again, I am stuck. How can I prove this inequality?

Another way.
Afte using your $uvw$'s substitution we see that our inequality is a linear inequality of $w^3$,
which by $uvw$ says that it's enough to prove our inequality in the following cases.


*

*$w^3\rightarrow0^+$.


Let $c\rightarrow0^+$ and $b=1$.
We obtain:
$$a(a+1)(a-1)^2\geq0;$$
2. Two variables are equal.
Let $b=c=1$.
We obtain:
$$a^2(a-1)^2\geq0.$$
About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791
