Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $

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.

1. $w^3\rightarrow0^+$.

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?