<s>[INST] 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})$?
 [/INST] 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.
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