Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$.
That is $\lVert uu^T-vv^T\rVert^2\_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.


Is there a one-liner proof?


Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?


Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15):
<https://arxiv.org/abs/1606.05225>
The eigenvalue of $||uu^T-vv^T||\_2$ seems to be correct by
[How to find the eigenvalues of $xx^T-yy^T$](https://math.stackexchange.com/questions/3913336/how-to-find-the-eigenvalues-of-x-xt-y-yt)


Edit 2: I assumed that all the conjectures in the question are equivalent but maybe I was wrong. I took the Schur Decomposition $USU^T$ of $uu^T-vv^T$ to get $\max\_{||x||=1} ||(uu^T-vv^T)x||^2=||USU^Tx||^2=S\_{1,1}^2$ and assumed it is the same $\max\_{||x||=1} |x^T(uu^T-vv^T)x|^2=|x^TUSU^Tx|=S\_{1,1}^2$. Then I noticed that $|x^T(uu^T-vv^T)x|^2=|(x^Tu)^2-(v^Tx)|$. Not sure what went wrong.


Summary: As GH from Mo noted below, I forgot a squared root in the right hand side and the statement is wrong. Hope to get your help also in the fixed version [here](https://mathoverflow.net/questions/417022/prove-that-vtx2%E2%88%92utx2-leq-sqrt1%E2%88%92utv2-for-any-unit-vectors-u-v)



Prove that $(v^Tx)^2-(u^Tx)^2 < 1-(u^Tv)^2$ for any unit vectors $u$, $v$, $x$
The inequality as stated is false, but it is true that
$$\langle v,x\rangle^2+\langle v,u\rangle^2\leq 1+|\langle x,u\rangle|.$$
Moreover, the right-hand side is optimal in the sense that it is the maximum of the left-hand side over the unit vectors $v$. More generally, if $u\_1,\dots,u\_R$ are any vectors in a Hilbert space, then
$$\max\_{\langle v,v\rangle=1}\sum\_{r=1}^R|\langle v,u\_r\rangle|^2$$
equals the largest eigenvalue of the Gram matrix $(\langle u\_s,u\_t\rangle)\_{1\leq s,t\leq R}$. For details, see [these notes](https://users.renyi.hu/%7Egharcos/bombieri-halasz-montgomery.pdf) on the Bombieri-Halász-Montgomery inequality.


