<s>[INST] 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! 
 [/INST] 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.
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