Have I done my homogenization correctly for this question and if so how do I finish it off?

Question

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?

Answer 1

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 $a^2+b^2+c^2=1$

Answer 2

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 $a^2+b^2+c^2=1$.