As Can Hang points out in his response, the inequality does not hold in general. Thanks to his comment to my own post, but I stand corrected and claim the inequality is valid at least for the case $a_2b_2\ge0$.$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 $$and note that A,B,C are nonnegativeand ; the inequality A\ge0 is equivalent to the above condition (*). In addition,$$ 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, $$while the latter follows from (2). 2 added 100 characters in body The inequality does not hold in general, but at least for the case a_2b_2\ge0. Let me do some standard things. First let$$ 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}. 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 $$and note that A,B,C are nonnegative and$$ 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, $$while the latter follows from (2). 1 Let me do some standard things. First let$$ 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}. $$Substitution reduces the inequality to the following one:$$ ((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 $$and note that A,B,C are nonnegative and$$ 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).