Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$

Question

For $a,b,c \in (0,1)$ such that $ab+bc+ca=1$ Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$

I tried repleace $1$ by $ab+bc+ca$ so we need to prove: $\frac{a^2+b^2}{(ab+bc+ca-a^2)(ab+bc+ca-b^2)} + \frac{b^2+c^2}{(ab+bc+ca-b^2)(ab+bc+ca-c^2)}+\frac{c^2+a^2}{(ab+bc+ca-c^2)(ab+bc+ca-a^2)} \geq \frac{9}{2}$

Then i try to factor the denominator. This didn't seem to help me.

Answer 1

Another way: $$\sum_{cyc}\frac{a^2+b^2}{(1-a)^2(1-b^2)}\geq\sum_{cyc}\frac{\frac{1}{2}(a+b)^2}{(1-ab)^2}=\sum_{cyc}\frac{\frac{1}{2}(a+b)^2}{c^2(a+b)^2}=$$ $$=\frac{1}{2}\sum_{cyc}\frac{1}{a^2}\geq \frac{1}{2}\sum_{cyc}\frac{1}{ab} =\frac{1}{2}\sum_{cyc}ab\sum_{cyc}\frac{1}{ab}\geq\frac{9}{2}.$$

Answer 2

By C-S $$\sum_{cyc}\frac{a^2+b^2}{(1-a^2)(1-b^2)}\geq\frac{2(a+b+c)^2}{\sum\limits_{cyc}(1-a^2)(1-b^2)}$$ and it's enough to prove that $$4(a+b+c)^2\geq9\sum_{cyc}(1-a^2)(1-b^2),$$ which after homogenization gives: $$\sum_{cyc}(11a^3b+11a^3c-14a^2b^2-8a^2bc)\geq0,$$ which is true by Muirhead.