I believe I found a complicated proof by bounding the spectral norm $||uu^T-vv^T||^2\_2:=\max\_{||x||=1}|(u^Tx)^2-(v^Tx)^2|$.
Using the fact that $dist(x,y):=\sin|x-y|$ is a distance function over unit vectors, we may prove
$$(v^Tx)^2-(u^Tx)^2=\sin^2(u,x)-\sin^2(v,x)=dist^2(u,x)-dist^2(v,x)
=(dist(u,x)-dist(v,x))(dist(u,x)+dist(v,x)) \leq 2dist(u,v),$$
but the approximation factor is too large.
Can you improve the bound using small changes to the above, or find another simple proof using trigo or Taylor's approx?


Note: This is a follow-up on a previous question that was stated wrong [here](https://mathoverflow.net/questions/414965/prove-that-vtx2-utx2-1-utv2-for-any-unit-vectors-u-v-x). I hope it is correct this time.



Prove that $(v^Tx)^2−(u^Tx)^2\leq \sqrt{1−(u^Tv)^2}$ for any unit vectors $u, v, x$
This is a variation of Fedor Petrov's proof (found independently). First, it suffices to prove the statement when $x$ lies in $V:=\mathbb{R}u+\mathbb{R}v$. Indeed, we see this reduction readily by decomposing $x$ as $cy+z$, where $c\in[0,1]$ is a scalar, $y\in V$ is a unit vector, and $z\in V^\perp$. By this reduction, we can assume that $V=\mathbb{R}^2$, and
$$\qquad x=(1,0),\qquad u=(\cos\alpha,\sin\alpha),\qquad v=(\cos\beta,\sin\beta).$$
So we need to prove that
$$\cos^2\beta-\cos^2\alpha\leq|\sin(\alpha-\beta)|.$$
However, this is clear, because the left-hand side equals $\sin(\alpha+\beta)\sin(\alpha-\beta)$.


**P.S.** I upvoted the question and the other three answers, for fun and world peace.


