Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$ It's an inequality based on two found on the website MSE (see the reference):
Let $a,b,c>0$ then we have:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Lemma 1 :
$a,b,c>0$ then we have :
$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Proof of lemma 1 :
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 :
$\left(\frac{((3u)((3u)^2-4(3v^2))+5w^3)}{3u3v^2-w^3}+2\right)^2\geq \frac{9}{4}+\frac{(9/4)((9u^2-6v^2)^2-2(9v^4-6uw^3)-(9v^4-6uw^3))+(2.25u+0.75w)(9u^2-9v^2)(3u)}{(2.25u+0.75w)(3u3v^2-w^3)}$
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 $b=c=1$.
$$\frac{2}{a+1}+\frac{a}{2}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(a^4+1-2a^2)}{((a+2)\frac{3}{4}+\frac{3}{4}(a)^{\frac{1}{3}})(a+1)^2(2)}+\frac{(a^2+1-2a)(a+2)}{(a+1)^2(2)}}$$
Now it seems to be clear : we get a polynomial  with a root equal to one . See the factorization by Wolfram alpha .
End of the proof of the lemma 1
Remains to show that $ a\geq b \geq c>0$:
$$P(a,b,c)\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Wich is not hard I think .
Question :
How to show it ?
Reference :
M. A. Rozenberg, “uvw–Method in Proving Inequalities”, Math. Ed., 2011, no. 3-4(59-60), 6–14
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
Stronger than Nesbitt inequality
 A: As username says it's not symmetric so here there is a refinement and a symmetric formula :
$a,b,c\in[0.95,1]$:
$$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(0.75(abc)^{\frac{1}{3}}+0.75(a+b+c))(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}+\frac{9}{4}\left(abc\frac{\frac{a^2+b^2+c^2}{ab+bc+ca}-1}{a^3+b^3+c^3}\right)^2}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
A: The given inequality is homogenious by $\;(a,b,c)\;$ and symmetric by $\;(a,b).\;$
Let WLOG
$$a+b+c = 6,\quad a+b=s,\quad ab=p,\tag1$$
then it suffices to prove the inequality in the form of
$$\dfrac a{6-a}+\dfrac b{6-b}+\dfrac{6-a-b}{a+b} \ge \sqrt{\dfrac94
+9\,\dfrac{(a-b)^2}{(a+b)(6-a)(6-b)}},\tag2$$
with the both positive sides, wherein the square root function monotonically increases in $\;[0,\infty).\;$
Therefore, one may to square both of the inequality sides and eliminate the positive denominators, with the represntations
\begin{align}
&\bigg(\big(a(6-b)+b(6-a)\big)(a+b)+(6-a-b)(6-a)(6-b)\bigg)^2\\[4pt]
&\ge \dfrac94\bigg(\big((6-a)(6-b)(a+b)\big)^2-4(a-b)^2(6-a)(6-b)(a+b)\bigg),
\tag3 \end{align}
or
$$\dfrac49\big(6s^2-2ps + 6(6-s)^2+p(6-s)\big)^2$$
$$\ge (36s-6s^2+ps)^2-4(s^2-4p)(36s-6s^2+ps),$$
or
$$f(s,p)\ge 0,\tag4$$
where
$$s^2\ge4p,\quad 6\ge s,\tag5$$
$$f(s,p)=\big(144-48s+8s^2+4p-2ps\big)^2 - (36s-8s^2+8p+ps)^2+4(s^2-4p)^2$$
$$=\big(144-84s+16s^2-4p-3ps\big)\big(144-12s+12p-ps\big)+4(s^2-4p)^2$$
$$\ge\dfrac34\big(192-112s+20s^2-s^3\big)(12-s)(12+p)$$
$$=\dfrac34(s-4)^2(12-s)^2(12+p)\ge0.$$
Proved!
