Prove or disprove : $ a\geq b\geq c\geq 1$:
$$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(3(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Motivation :
As linked in the comment for @Albus Dumbledore it's an attempt to make the HN_NH's problem more symmetric and use $uvw's$ method .Again it's not a messy result because I have spend some hours to find it .
First we remark that the inequality is homogenous and we can try the substitution $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$ and apply the uvw's method .
We have : $$a^4+b^4+c^4=(9u^2-6v^2)^2-2(9v^4-6uw^3)$$ $$a^2b^2+b^2c^2+c^2a^2=9v^4-6uw^3$$ $$a^2+b^2+c^2=9u^2-6v^2$$
And : $\left(\frac{((3u)((3u)^2-4(3v^2))+5w^3)}{3u3v^2-w^3}+2\right)^2\geq \frac{9}{4}+\frac{(9/4)((9u^2-6v^2)^2-2(9v^4-6uw^3)-(9v^4-6uw^3))+3w(9u^2-9v^2)(3u)}{(3w)(3u3v^2-w^3)}$
it's enough to find an extreme value of our expression for the extreme value of $w^3$ wich happens for an equality case of two variables .
Since the last inequality is homogeneous, we can assume that $b=c=1$.
$$\frac{2}{a+1}+\frac{a}{2}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(a^4+1-2a^2)}{(3(a)^{\frac{1}{3}})(a+1)^2(2)}+\frac{(a^2+1-2a)(a+2)}{(a+1)^2(2)}}$$
Now it seems to be clear : we get a long polynomial with a root equal to one . See the factorization by Wolfram alpha
But I think that it misses something in my proof.
If there is a mistake feel free to comment.
Is my proof correct ? How to show it with another method?
Ps:One downvote a day, keeps the truth away.
PPS:Don't forget the constraint above !
