An Inequality Problem $1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$ If $a,b,c>0$, are positive real numbers such that $a^2+b^2+c^2=1$ then, the following Inequalities hold:
$\displaystyle 1 \le \frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ac} \le \frac{3\sqrt{3}}{2}$
$\displaystyle 1 \le \frac{a}{1+ab}+\frac{b}{1+bc}+\frac{c}{1+ac} \le \frac{3\sqrt{3}}{4}$
Homogenizing the first inequality as, 
$\displaystyle \frac{a}{a^2+b^2+c^2-ab}+\frac{b}{a^2+b^2+c^2-bc}+\frac{c}{a^2+b^2+c^2-ac} \le \frac{3\sqrt{3}}{2\sqrt{a^2+b^2+c^2}}$
and noting the cyclic symmtery I tried Rearrangement Inequalities Inequalities.
Using $a^2+b^2+c^2-ab = \frac{1}{4}(a+b)^2+\frac{3}{4}(a-b)^2+c^2 \ge \frac{1}{4}(a+b)^2+c^2 \ge (a+b)c$
$\displaystyle \sum_{cyc} \frac{a}{a^2+b^2+c^2-ab} \le \sum_{cyc} \frac{a}{ac+bc}$ might help, but I couldn't get anywhere with it.
Thank you!

Please consider the following as a comment, I'm not able to say if the Holder equality is valid in that case.  

RHS for the first inequality :
$$1-ab=1+\frac{(a-b)^2-a^2-b^2}{2}=\frac12+\frac{(a-b)^2+c^2}{2}\geq \frac{1+c^2}{2}$$
so
$$\sum_{cyc} \frac{a}{1-ab}\leq \sum_{cyc} \frac{a}{\frac{1+c^2}{2}}$$
We then use the Hölder's inequality :
$$\sum_{cyc} \frac{a}{1-ab}\leq \left(\sum_{cyc} a^2\right)^{\frac12}\left(\sum_{cyc} \frac{1+c^2}{2}\right)^{-1}\left(\sum_{cyc} 1\right)^{\frac32}=\frac{3\sqrt{3}}{2}$$
LHS for the first inequality :
$$1-ab\leq 1$$
so
$$\sum_{cyc} \frac{a}{1-ab}\geq \sum_{cyc} a$$
Or :
$$1=\sum_{cyc} a^2\leq \left(\sum_{cyc} a\right)^{2}$$
so
$$1\leq \sum_{cyc} \frac{a}{1-ab}$$
