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 $m\geq {9\over 8}$.
 A: Your calculation is correct: $m = \frac 65$ is an upper bound. It is not the best upper bound though, because the estimates $\frac{2}{5+c^2} < \frac 25$ are not sharp for all three variables simultaneously.
The best upper bound is $\color{red}{m=\frac 98}$. Proof: The function $f(t) = \frac{1}{3-\sqrt t}$ is increasing and concave on $[0, 1]$:
$$
f'(t) = \frac{1}{2 \sqrt t (3-\sqrt t)^2} \ge 0 \, ,\\
f''(t) = -\frac{3(1-\sqrt t)}{4 t^{3/2}(3-\sqrt t)^3} \le 0 \, .
$$
Therefore Jensen's inequality gives
$$
\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} = f(a^2b^2)+f(b^2c^2)+ f(c^2a^2) \\
\le 3 f\left( \frac{a^2b^2+b^2c^2+c^2a^2}{3}\right)
 \underset{(*)}{\le} 3 f\left( \bigl( \frac{a^2+b^2+c^2}{3} \bigr)^2\right) 
= 3 f(\frac 19) = \frac 98 \, .
$$
At $(*)$ we have used that
$$
 \frac{a^2b^2+b^2c^2+c^2a^2}{3} \le \left( \frac{a^2+b^2+c^2}{3} \right)^2 \, ,
$$
which is MacLaurin's inequality applied to
$$
 (x_1, x_2,x_3) = (a^2, b^2, c^2) \, .
$$
So $m= \frac98$ is an upper bound, and it is best possible since equality holds for $a=b=c=1/\sqrt 3$.
A: 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.
