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! 

My second proof:
Fact 1: $\frac{1}{(1-x)^2} \le \frac{729}{40}x^4 + \frac{243}{40}x^2 + \frac{27}{20}$ for all $x\in [0, 1/2]$.
(Note: $\frac{729}{40}x^4 + \frac{243}{40}x^2 + \frac{27}{20} - \frac{1}{(1-x)^2}
= \frac{(81x^4 - 108x^3 + 27x^2 - 24x + 14)(1-3x)^2}{40(1-x)^2} \ge 0$.)
Noting that $ab, bc, ca \le 1/2$, by Fact 1, it suffices to prove that
$$\frac{729}{40}((ab)^4 + (bc)^4 + (ca)^4) + \frac{243}{40}((ab)^2 + (bc)^2 + (ca)^2) + \frac{81}{20}
\le \frac{27}{4}$$
or
$$27(a^4b^4 + b^4c^4 + c^4a^4) + 9(a^2b^2 + b^2c^2 + c^2a^2) \le 4.$$
Letting $x = a^2, y = b^2, z = c^2$,
we have $x + y + z = 1$. We need to prove that
$$27(x^2y^2 + y^2z^2 + z^2x^2) + 9(xy + yz + zx) \le 4. \tag{1}$$
We use pqr method.
Let $p = x + y + z = 1, q = xy + yz + zx, r = xyz$. Using $p^2 \ge 3q$, we have $q \le 1/3$.
The desired inequality (1) is written as
$$27(q^2 - 2pr) + 9q \le 4.$$
We split into two cases.
Case 1: $1/4 < q \le 1/3$
Degree three Schur yields
$r \ge \frac{4pq - p^3}{9} = \frac{4q - 1}{9}$. We have
$$27(q^2 - 2pr) + 9q - 4
\le 27\left(q^2 - 2p\cdot \frac{4q - 1}{9}\right) + 9q - 4
= (9q - 2)(3q - 1) \le 0.$$
Case 2: $0 \le q \le 1/4$
We have
$$27(q^2 - 2pr) + 9q - 4 \le 27q^2 + 9q - 4 < 0.$$
We are done.
