Another plausible inequality. I come across the following problem in my study. 
Consider in the real field. Let $ 0\le x\le1 $, $a_1^2+a_2^2=b_1^2+b_2^2=1$.Is it true
$ (a_1b_1+xa_2b_2)^2\le\left(\frac{(1-x)+(1+x)(a_1b_1+a_2b_2)}{(1+x)+(1-x)(a_1b_1+a_2b_2)}\right)^{2}(a_1^2+xa_{2}^{2})(b_1^2+xb_{2}^{2})$?
 A: I think your inequality is false, dear miwalin. Please check the case when $a_1=b_2=\frac{\sqrt{3}}{2}$ and $a_2=b_1=-\frac{1}{2}.$ But I think it is true when $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative numbers.
Let me prove it in the case $a_1,$ $a_2,$ $b_1,$ $b_2$ are nonnegative real numbers. Write the inequality as
$$\frac{(a_1^2+xa_2^2)(b_1^2+xb_2^2)}{(a_1b_1+xa_2b_2)^2} -1 \ge \left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1.$$
Since 
$$(a_1^2+xa_2^2)(b_1^2+xb_2^2)-(a_1b_1+xa_2b_2)^2=x(a_1^2b_2^2+a_2^2b_1^2-2a_1a_2b_1b_2)= x[(a_1^2+a_2^2)(b_1^2+b_2^2)-(a_1b_1+a_2b_2)^2]= x[1-(a_1b_1+a_2b_2)^2]$$
and
$$\left[ \frac{(1+x)+(1-x)(a_1b_1+a_2b_2)}{(1-x)+(1+x)(a_1b_1+a_2b_2)}\right]^2-1=\frac{4x[1-(a_1b_1+a_2b_2)^2]}{[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2},$$
the above inequality is equivalent to (notice that $x[1-(a_1b_1+a_2b_2)^2] \ge 0$)
$$[(1-x)+(1+x)(a_1b_1+a_2b_2)]^2 \ge 4(a_1b_1+xa_2b_2)^2,$$
or
$$(1-x)+(1+x)(a_1b_1+a_2b_2) \ge 2(a_1b_1+xa_2b_2),$$
or
$$(1-x)(1-a_1b_1+a_2b_2) \ge 0,$$
which is obvious.
A: 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).
