<s>[INST] 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!
 [/INST] The left inequalities we can prove by Holder:


*

*$$\left(\sum_{cyc}\frac{a}{1-ab}\right)^2\sum_{cyc}a(1-ab)^2\geq(a+b+c)^3.$$


Thus, it remains to prove that
$$(a+b+c)^3\geq \sum_{cyc}a(1-ab)^2$$ or
$$(a+b+c)^3\geq(a+b+c)(a^2+b^2+c^2)-2\sum_{cyc}a^2b+\sum_{cyc}a^3b^2$$ or
$$\sum_{cyc}(4a^2b+2a^2c+2abc-a^3b^2)\geq0,$$ 
for which it's enough to prove that
$$\sum_{cyc}a^2c\geq\sum_{cyc}a^3b^2$$ or
$$\sum_{cyc}ab^2\sum_{cyc}a^2\geq\sum_{cyc}a^3b^2,$$
which is obvious;


*$$\left(\sum_{cyc}\frac{a}{1+ab}\right)^2\sum_{cyc}a(1+ab)^2\geq(a+b+c)^3.$$


Thus, it remains to prove that
$$(a+b+c)^3\geq \sum_{cyc}a(1+ab)^2$$ or
$$(a+b+c)^3\geq(a+b+c)(a^2+b^2+c^2)+2\sum_{cyc}a^2b+\sum_{cyc}a^3b^2$$ or
$$\sum_{cyc}(2a^2c+2abc-a^3b^2)\geq0,$$ 
for which it's enough to prove that
$$\sum_{cyc}a^2c\geq\sum_{cyc}a^3b^2,$$
which is obvious.
The right inequalities are true, but I still have no a nice proof.
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