Well,I think you can use the famouse Iran 96 inequality as a result to solve this problem.
the Iran 96 inequality
Let $x,y,z\geq 0$ we have
$$ \frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(x+z)^{2}}\geq \frac{9}{4(xy+yz+zx)} $$
square both side,we can rewrite the inequality into
$$ \sum_{cyc}{\frac{1}{(x+y)^{2}}}+2\sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq \frac{25}{4}$$
Now,Using this inequality as a known result,it's suffice to prove
$$ \sum_{cyc}{\frac{1}{(x+y)(x+z)}}\geq 2 $$
after simple homogenous,it's
$$ (xy+yz+xz)(x+y+z)\geq (x+y)(y+z)(z+x) $$
Or
$$ xyz\geq 0 $$
Which is obviously true,Equality occurs if and only if $x=y=1, z=0 $ or it's permutation.
The proof is complete
Your first step gives a wrong inequality. Try $c\rightarrow0+$ and $a=b=1$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\frac{\sum\limits_{cyc}a^2(a+b)(a+c)}{9uv^2-w^3}+\frac{12}{u}\geq20.$$
Now, we see that it's a linear inequality of $w^3$ because $\sum\limits_{cyc}a^2(a+b)(a+c)$ is fourth degree and the condition does not depend on $w^3$.
Indeed, $$\sum_{cyc}a^2(a+b)(a+c)=\sum_{cyc}a^2(a(a+b+c)+bc)=$$
$$=(a^3+b^3+c^3)(a+b+c)+(a+b+c)abc=(27u^3-27uv^2+3w^3)3u+3uw^3.$$
Hence, we need to prove that
$$\frac{(27u^3-27uv^2+3w^3)3u+3uw^3}{9uv^2-w^3}+\frac{12}{u}\geq20,$$ which is a linear inequality of $w^3$ after full expanding.
Thus, it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.
- $w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
Thus, we need to prove that
$$\frac{a^2}{\frac{1}{a}}+\frac{\frac{1}{a^2}}{a}+\frac{36}{a+\frac{1}{a}}\geq20$$ or
$$(a-1)^4(a^4+4a^3+11a^2+4a+1)\geq0;$$
2. Two variables are equal.
Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a<1$.
Thus, we need to prove that:
$$\frac{2a^2}{a+\frac{1-a^2}{2a}}+\frac{\left(\frac{1-a^2}{2a}\right)^2}{2a}+\frac{36}{2a+\frac{1-a^2}{2a}}\geq20$$ or
$$(1-a)(1+a+3a^2-157a^3+415a^4-225a^5+381a^6-99a^7)\geq0,$$
which is smooth.
We can use also the following way.
$$\sum_{cyc}\frac{a^2}{b+c}=\sum_{cyc}\frac{a^2(ab+ac+bc)}{b+c}=a^3+b^3+c^3+\sum_{cyc}\frac{a^2bc}{b+c}=$$
$$=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc+\sum_{cyc}\frac{a^2bc}{b+c}\geq$$
$$\geq(a+b+c)^3-3(a+b+c).$$
Id est, it's enough to prove that
$$(a+b+c)^3-3(a+b+c)+\frac{36}{a+b+c}\geq20.$$
Can you end it now?
Best Answer
Using the Cauchy-Scwarz inequality
$$ (a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}) \geq (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^{2} $$ with $a_{1}=\frac{x+y-1}{\sqrt{z}} $, $ a_{2}=\frac{x+z-1}{\sqrt{y}} $ $ a_{3}=\frac{z+y-1}{\sqrt{x}} $ and $ b_{1}=\sqrt{z} $, $ b_{2}=\sqrt{y} $ and $ b_{3}=\sqrt{x} $, we obtain that $$\Big (\frac{(x+y-1)^{2}}{z}+\frac{(x+z-1)^{2}}{y}+\frac{(z+y-1)^{2}}{x} \Big) \cdot S \geq (2S-3)^{2} $$ where $ S=x+y+z $. Therefore it suffices to show that $\frac{(2S-3)^{2}}{S} \geq 12 $ which is equivalent to showing that $ S \in (-\infty, \frac{6-3\sqrt{3}}{2}]\cup [ \frac{6+3\sqrt{3}}{2},\infty ) $.
But we have that $ S=x+y+\frac{8}{xy} $ and using the AM-GM, this is greater than or equal to $ 3(xy\cdot 8\frac{1}{xy})^{\frac{1}{3}}=6 $.
since clearly $ 6> \frac{6+3\sqrt{3}}{2} $, the conclusion follows.
Moreover, $ \frac{27}{2} $ is the sharpest lower bound and in the same manner we can prove that the LHS is $ \geq \frac{27}{2} $ since this is equivalent to $ S \in (-\infty, \frac{3}{8}] \cup [6, \infty) $ after computations and this is true since we proved earlier that $ S \geq 6 $ .