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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?

Your first step gives a wrong inequality. Try $crightarrow0+$ and $a=b=1$.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$frac{sumlimits_{cyc}a^2(a+b)(a+c)}{9uv^2-w^3}+frac{12}{u}geq20.$$
Now, we see that it's a linear inequality of $w^3$ because $sumlimits_{cyc}a^2(a+b)(a+c)$ 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. 
   $w^3rightarrow0^+$.

Let $crightarrow0^+$.

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?