If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$ 
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
  $$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$

This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with the same condition, 
which we can prove by AM-GM and C-S:
$$\sum\limits_{cyc}\frac{1}{1-ab}=3+\sum\limits_{cyc}\left(\frac{1}{1-ab}-1\right)=3+\sum\limits_{cyc}\frac{ab}{1-ab}\leq$$
$$\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(2a^2+2b^2+2c^2-2ab)}\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(a^2+b^2+2c^2)}\leq$$
$$\leq3+\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2},$$
but for the starting inequality this idea does not work.
By the way, I have a proof of the following inequality.
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=3$. Prove that:
 $$\sum\limits_{cyc}\frac{1}{(4-ab)^2}\leq\frac{1}{3}$$
(we can prove it by SOS and uvw). 
This inequality is weaker and it not comforting.
We can assume of course that all variables are non-negatives.
Thank you! 

Let $x = \dfrac{1}{1-ab}$, $y = \dfrac{1}{1-bc}$, $z = \dfrac{1}{1-ac}$.
We have proven that
$$x + y + z \leq \dfrac{9}{2}.$$
Therefore, $x + y + z = \dfrac{9}{2} - \varepsilon$ for some $0 \leq \varepsilon \leq \dfrac{9}{2}$. If we assume that $x,y,z \geq 0$, which we can, we can square both sides and obtain
$$(x^2 + y^2 + z^2) + (2xy + 2yz + 2xz) = \bigg{(} \dfrac{9}{2} - \varepsilon \bigg{)}^2 \leq \dfrac{81}{4}$$
We can say that $x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2(xy + yz + xz)$. We now seek to minimize $xy + yz + xz$ in order to maximize the upper bound.
We are given that $a^2 + b^2 + c^2 = 1$, and since $xy + yz + xz$ is a symmetric polynomial in $x,y,z$, we should seek $x=y=z>0$ as our minimum, which can be shown the be the case using very straightforward techniques from calculus (Second derivative test should suffice, proving the minimum for $a,b,c$ suffices). We find that $x=y=z$ when $a=b=c=\dfrac{1}{\sqrt{3}}$, so that $x=y=z=\dfrac{1}{1-1/3}=\dfrac{3}{2}$. We can then say that
$$xy + yz + xz \geq 3*(3/2)^2 = 3*(9/4) = 27/4$$
Therefore,
$$x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2*\dfrac{27}{4} = \dfrac{81}{4} - \dfrac{54}{4} = \dfrac{27}{4}$$
