Have I done my homogenization correctly for this question and if so how do I finish it off? We have $a,b,c>0$ with $a^2+b^2+c^2=1$ prove that:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>1$
As soon as I saw this question, I immediately thought of using homogenization in the following manner:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>\frac{1}{\sqrt{a^2+b^2+c^2}}$
Then I tried using Andreescu:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{(1+bc)(1+ac)(1+ab)}$
Which didn't work out.
Could you please explain to me if I have done my homogenization correctly? If I have done it correctly, could you please show me how to finish it off and if I haven't how to think of doing the correct homogenization?

we will prove $$\sum_{cyc}\frac{a}{1+bc}=3-\sum_{cyc}\frac{abc}{1+bc}>1$$
or $$\sum\frac{abc}{1+bc}<2$$
we know will make denominator homogenous in degree 2
$1+bc=a^2+b^2+c^2+bc\ge 4a^{1/2}b^{3/4}c^{3/4}$
thus  $$\sum\frac{abc}{1+bc}\le \frac{1}{4}a^{1/2}b^{1/4}c^{1/4}\le \frac{3}{4}abc\le \frac{1}{4\sqrt{3}}<2$$\
here we used $$abc\le \frac{1}{3\sqrt{3}}$$ by am-gm using $a^2+b^2+c^2=1$
