As Can Hang points out in his response, the inequality does not hold in general. Thanks to his comment to my own post, I stand corrected and claim the inequality is valid at least for the case
$$
a_1b_1+xa_2b_2\ge0
\qquad(*)
$$
(and this seems to be a necessary condition as well).

Let me do some standard things. First let
$$
a_1=\frac{1-u^2}{1+u^2}, \quad
a_2=\frac{2u}{1+u^2}, \quad
b_1=\frac{1-v^2}{1+v^2}, \quad
b_2=\frac{2v}{1+v^2}
$$
where $uv\ge0$.
Substitution reduces the inequality to the following one:
$$
((1-u^2)(1-v^2)+4xuv)^2
\le\biggl(\frac{(uv+1)^2-x(u-v)^2}{(uv+1)^2+x(u-v)^2}\biggr)^2
((1-u^2)^2+4xu^2)((1-v^2)^2+4xv^2).
\qquad{(1)}
$$
Now introduce the notation
$$
A=(1-u^2)(1-v^2)+4xuv, \quad
B=(uv+1)^2, \quad C=x(u-v)^2
$$
and note that $A,B,C$ are nonnegative; the inequality $A\ge0$ is equivalent to the above condition $(*)$. In addition,
$$
A\le B-C
\qquad{(2)}
$$
because
$$
B-C-A=(1-x)(u+v)^2\ge0.
$$
In the new notation the inequality (1) can be written more compact:
$$
A^2(B+C)^2\le(B-C)^2(A^2+4BC)
$$
which after straightforward reduction becomes
$$
A^2\le(B-C)^2,
$$
while the latter follows from (2).