<s>[INST] Prove $\sum_\text{cyc}\frac{1}{3-ab}\le\frac{3}{2}$ 
Let $a,b,c>0$ such that $a^2+b^2+c^2=1$. Prove that $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le\frac{3}{2}$$

By Am/Gm we have: $$ {1\over 3-ab}\leq {2\over 6-a^2-b^2} = {2\over 5+c^2}<{2
\over 5}$$
so 
$$\sum_{cyc} {2\over 5+a^2} < {6\over 5} <{3\over 2}$$
So I get much better bound. Where did I go wrong?

Edit: If nothing is wrong then what is a smallest constant $m$ such  $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le m$$ For $a=b=c=\sqrt{3}/3$ we get $m\geq {9\over 8}$.
 [/INST] Hint.-Because of $a^2+b^2+c^2=1$, each $a,b,c$ must be in the interval $[-1,1]$ then one has for positive values
$$\frac13\le\frac{1}{3-ab}\le\frac12\\\frac13\le\frac{1}{3-ac}\le\frac12\\\frac13\le\frac{1}{3-bc}\le\frac12$$ Thus the sum is less or equal than $\dfrac32$.
If some of $a,b,c$ is negative, the fraction is less than $\dfrac13$ then one has the same majorant $\dfrac32$ for the sum.
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