Prove: $\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge4$ 
Let $a,b,c\ge0: (ab)^2+(bc)^2+(ca)^2=1.$ Prove that: $$\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge4$$

My approach using well- known inequality:
For all $a,b,c\ge0$ then: $$\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\ge\frac{3}{ab+bc+ca}$$ It implies that we need to prove: $$\sqrt{a^2b^2+b^2c^2+c^2a^2}\frac{3}{ab+bc+ca}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge4$$ But this one is not true for $(a,b,c)=(0,9;0,5;\dfrac{11}{28})$
I hope we can find a good way to solve the problem. Thanks!

Now, I have a partly solution due to my teacher's hint.Using the following one:
$$\frac{a}{a^{2}+bc}\ge\frac{3ab+3ca-a^{2}-bc}{(a+b)(b+c)(c+a)}$$$$\iff (a^2+bc -ab-ac)^2 \geq 0$$
$$\implies \frac{a^2}{a^{2}+bc}\ge\frac{3a^2b+3ca^2-a^{3}-abc}{(a+b)(b+c)(c+a)}$$ Hence $$ \frac{a^2}{a^{2}+bc}+ \frac{b^2}{b^{2}+ca}+ \frac{c^2}{c^{2}+ab}+\frac{a^3+b^3+c^3+9abc}{(a+b)(b+c)(c+a)}\ge3$$
Now, we need to prove that: $$\frac{1-a^2}{a^2+bc}+\frac{1-b^2}{b^2+ca}+\frac{1-c^2}{c^2+ab}\ge1 (*)$$
But how to prove this nice one and using the original condition?
Edit: (*) is equivalent to: $$\sum_{cyc}{\frac{a^2(b^2+c^2)\left(b^2+c^2-a^2-2bc\right)^2}{(a^2+bc)\left(a^2+b^2+c^2+(b^2+c^2-a^2)bc+2a^2(b^2+c^2)\right)}}\ge0$$
I see a similar problem on AOPS which help very well. Also, I hope to see more approach for (*).
