Lemma. Let $a, b, c, d, e$ and $f$ be real numbers. If for all real numbers $\lambda$ and $\mu$, we have
$$a + b\lambda^2 + c\mu^2 + 2(d\lambda + e\lambda\mu + f\mu) \ge 0, \qquad (1)$$
then we have
$$abc + 2def \ge ae^2 + bf^2 + cd^2. \qquad (2)$$

Proof. Clearly, $b \ge 0$. If $b=0$, then it is easy to see that $d=e=0$, thus (2) holds. Thus we may assume $b > 0$.
Since the left-hand side of (1) is quadratic in the variable $\lambda$, its discriminant is non-positive, i.e., $(d+e\mu)^2 - b(a+2f\mu+c\mu^2) \le 0$. Hence
$$(bc - e^2)\mu^2 + 2(bf - de)\mu + (ab - d^2) \ge 0.$$

Again, its left-hand side is quadratic in $\mu$ and so its discriminant is non-positive:
$$b \cdot (abc + 2def - ae^2 - bf^2 - cd^2) \ge 0.$$
This proves the claim.

By the Lemma, it suffices to show that for any real numbers $\lambda$ and $\mu$, we have
$$\frac{1}{1-x^2} + \frac{\lambda^2}{1-y^2} + \frac{\mu^2}{1-z^2} + 2 \left( \frac{\lambda}{1-xy} + \frac{\lambda\mu}{1-yz} + \frac{\mu}{1-zx} \right) \ge 0.$$
Using the summation formula of geometric series we obtain that the left side of the inequality equals
$$\sum_{n=0}^{\infty} (x^n + \lambda y^n + \mu z^n)^2$$
which is clearly non-negative.

The equality holds if and only if there exist real numbers $\lambda, \mu$ such that $x^n + \lambda y^n + \mu z^n = 0$ for any $n \ge 0$. So the equality holds if and only if two of $x, y, z$ are equal.