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The

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).
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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).
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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).