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?
 A: 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?
A: I can easy get an SOS (Sum of Squares)!
We have: $$\text{LHS-RHS} =\frac{A}{(a+b)(b+c)(c+a)(a+b+c)}\geqq 0$$
Where: $$A=\Big[4\, \left( a+b+c+5+\sqrt {3} \right)  \left( a+b+c-\sqrt {3} \right) 
+20\,\sqrt {3}-24\Big]abc$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b+c \right) \Big[ \left( a+b+c \right) ^{2
}+4\,a+4\,b+4\,c+9\Big] \left( a+b+c-2 \right) ^{2} \geqq 0$$
Equality holds when $a=b,\,c=0$ and its permutations.
A: 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.
