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?

Observe that $$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab}  = \sum_{cic} \frac{a^2}{a+abc} \geqslant \sum_{cyc} \frac{a^2}{a+a \cdot \frac{b^2+c^2}{2}}$$ Using $a^2+b^2+c^2=1$, we obtain  $$ \sum_{cyc} \frac{a^2}{a+a \cdot \frac{b^2+c^2}{2}}= \sum_{cyc} \frac{a^2}{a+a \cdot \frac{1-a^2}{2}} = \sum_{cyc} \frac{a^2}{1-\frac{(a+2)(a-1)^2}{2}}$$ This final sum is clearly smaller than $a^2+b^2+c^2=1$.
