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?

We need to prove that:
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4(ab+ac+bc)\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, $$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)=\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4-abc(a^2-2ab+b^2))=$$
$$=\sum_{cyc}(a-b)^2(ab(a+b)-abc)=\sum_{cyc}(a-b)^2ab(a+b-c)\geq$$
$$\geq(a-c)^2ac(a+c-b)+(b-c)^2bc(b+c-a)\geq$$
$$\geq(b-c)^2ac(a-b)+(b-c)^2bc(b-a)=(a-b)^2(b-c)^2c\geq0.$$
