Cho `{(a;b;c>0),(a^2+b^2+c^2=3):}`. CMR: `sum1/(3-ab)<=3/2`

$\dfrac{1}{3-ab}-\dfrac{1}{3}$ =\frac{1}{3}(\frac{ab}{3-ab})
$AM-GM:$
$ab \le \dfrac{a^2+b^2}{2}=\dfrac{3-c^2}{2}$
$=>$ $3-ab \ge 3-\dfrac{3-c^2}{2}=\dfrac{3+c^2}{2}=\dfrac{a^2+b^2+2c^2}{2}$
$=>$ $\dfrac{1}{3}(\dfrac{ab}{3-ab}) \mathop{\le}\limits^{AM-GM} \dfrac{1}{3}(\dfrac{\dfrac{(a+b)^2}{4}}{\dfrac{a^2+b^2+2c^2}{2}})=\dfrac{1}{6}(\dfrac{(a+b)^2}{a^2+b^2+c^2+c^2})$
$CBS:$
$\dfrac{1}{6}(\dfrac{(a+b)^2}{a^2+c^2+b^2+c^2}) \le\dfrac{1}{6}( \dfrac{a^2}{a^2+c^2}+\dfrac{b^2}{b^2+c^2})$
Tương tự: $\dfrac{1}{3-bc}-\dfrac{1}{3} \le \dfrac{1}{6}(\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{a^2+c^2})$
$\dfrac{1}{3-ac}-\dfrac{1}{3} \le \dfrac{1}{6}(\dfrac{a^2}{a^2+b^2}+\dfrac{c^2}{b^2+c^2})$
$=>$ $\sum \dfrac{1}{3-ab} – 1 \le \dfrac{1}{6}(\dfrac{a^2+b^2}{a^2+b^2}+\dfrac{b^2+c^2}{b^2+c^2}+\dfrac{a^2+c^2}{a^2+c^2})=\dfrac{1}{2}$
$=>$ $\sum \dfrac{1}{3-ab} \le 1+\dfrac{1}{2}=\dfrac{3}{2}$
Dấu $”=”:a=b=c=1$