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})$?



Another plausible inequality.
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).


