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!
 A: we can prove
$$\dfrac{x}{1+xy}\le\dfrac{3\sqrt{3}x(x+y+2z)}{4(x^2+y^2+z^2+3(xy+yz+xz))}$$
so
$$\sum_{cyc}\dfrac{x}{1+xy}\le \sum_{cyc}\left(\dfrac{3\sqrt{3}x(x+y+2z))}{4(x^2+y^2+z^2+3(xy+yz+xz)}\right)=\dfrac{3\sqrt{3}}{4}$$
I have an other ugly methods:
this problem is equivalent:$a^2+b^2+c^3=3$,then we have
$$\dfrac{a}{ab+3}+\dfrac{b}{bc+3}+\dfrac{c}{ca+3}\le\dfrac{3}{4}$$
$$\Longleftrightarrow 
4abc\sum_{cyc}ab+12\sum_{cyc}ab^2+36abc+36\sum_{cyc}a
\le 3a^2b^2c^2+9abc\sum_{cyc}a+27\sum_{cyc}ab+81$$
use this follow well know reslut
$$ab^2+bc^2+ca^2\le\dfrac{4}{27}(a+b+c)^3-abc$$
so we only
$$4abc\sum_{cyc}ab+12\left(\dfrac{4}{27}(a+b+c)^3-abc\right)
+36abc+36\sum_{cyc}a\le 3a^2b^2c^2+9abc\sum_{cyc}a+27\sum_{cyc}ab+81$$
let
$$p=a+b+c,q=ab+bc+ac,r=abc,p^2-2q=3$$
use AM-GM inequality we have
$$\dfrac{(p^2-3)^2}{4}=q^2\ge 3qr$$
so we consider
$$f(r)=3r^2-(2p^2-9p+18)r-\dfrac{16}{9}p^3+\dfrac{27}{2}p^2-36p
+\dfrac{81}{2}\ge 0$$
so
$$f'(r)=6r-2p^2+9p-18\le\dfrac{(p^2-3)^2}{2p}-2p^2+9p-18=
\dfrac{(p-1)(p-3)(p^2+2)-18}{2p}\le 0$$
so
$$f(r)\ge f\left(\dfrac{(p^2-3)^2}{12}\right)=\dfrac{(p-3)(3p^7-15p^6+27p^5-247p^4
+717p^3-1953p^2+621p-81)}{144p^2}$$
and we can prove 
$$3p^7-15p^6+27p^5-247p^4
+717p^3-1953p^2+621p-81\le 0(p\ge \sqrt{3})$$
A: 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
$$1\leq \sum_{cyc} \frac{a}{1-ab}$$
A: 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.
