Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $ Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$
What I have tried is firstly using the inequality: 
$\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, for any $x, y$ and any $a,b > 0$.
Using this inequality we obtain $\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} \geq \frac{a + b + c}{2}$, and then we have:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq \frac{a + b + c}{2} + \frac{36}{a + b + c} = \frac{(a + b + c)^2 + 72)}{2(a + b + c)} $$.
Using $ab + bc + ca = 1$, we would then have to prove that:
$$a^2 + b^2 + c^2 + 74 - 40(a + b + c) \geq 0 $$ and then I tried replacing in this inequality $c = \frac{1 - ab}{a + b}$, but I didn't get anything nice.
I also tried rewriting the lhs:
$$\frac{a^2}{b + c} = \frac{a^2(ab + bc + ca)}{b + c} = a^3 + \frac{a^2bc}{b + c}$$
And this would result in: $a^3 + b^3 + c^3 + abc(\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}) + \frac{36}{a + b + c} \geq 20$, but I didn't know how to continue from here.
Do you have any suggestions for this inequality?

Let $p=a+b+c,\,q=ab+bc+ca,$ we write your inequality as
$$ \sum \frac{a^2}{b+c} + \frac{36(ab+bc+ca)} {a+b+c}\geqslant 20\sqrt{ab+bc+ca},$$
or
$$(a+b+c) \sum \frac{a}{b+c} - (a+b+c) + \frac{36(ab+bc+ca)} {a+b+c}\geqslant 20\sqrt{ab+bc+ca}.$$
Use
$$ \sum \frac{a}{b+c} \ge \frac{a^2+b^2+c^2}{ab+bc+ca} = \frac{p^2}{q}-2.$$
We need to prove
$$p\left(\frac{p^2}{q}-2\right) - p + \frac{36q} {p}\geqslant 20\sqrt{q},$$
equivalent to
$$\frac{(p^4-3p^2q+36q^2)^2}{p^2q^2} \geqslant 400q,$$
or
$$\frac{(p^4+2p^2q+81q^2)(p^2-4q)^2}{p^2q^2} \geqslant 0.$$
Equality occur when $a=b=1,\,c=0.$ The proof is completed.
