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! 
 A: 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,
$$x^2 + y^2 + z^2 \leq \dfrac{81}{4} - 2*\dfrac{27}{4} = \dfrac{81}{4} - \dfrac{54}{4} = \dfrac{27}{4}$$
A: The Buffalo Way works. (I think that Michael Rozenberg knew this.)
We only need to prove the case $a, b, c > 0$.
After homogenization and clearing the denominators, it suffices to prove that 
$f(a, b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial of degree $12$.
Due to symmetry, assume that $c\le b\le a$. Let $c = 1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$.
Then $f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. We are done.
A: 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.
