Question 129
For real numbers $-1 \le a, b, c \le 1$ satisfying $a^2 + b^2 + c^2 = 2abc + 1$, prove
$$
\frac{3}{2 - abc} \le \frac{1}{2 - a^2} + \frac{1}{2 - b^2} + \frac{1}{2 - c^2} \le 1 + \frac{2}{2 - abc}.
$$
(Proposed by Otgonbayar Uuye)
$$
\frac{3}{2 - abc} \le \frac{1}{2 - a^2} + \frac{1}{2 - b^2} + \frac{1}{2 - c^2} \le 1 + \frac{2}{2 - abc}.
$$
(Proposed by Otgonbayar Uuye)