Prove that $(v^Tx)^2−(u^Tx)^2\leq \sqrt{1−(u^Tv)^2}$ for any unit vectors $u, v, x$ 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. I hope it is correct this time.
 A: 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.
A: As is observed, since
$$|\langle v,x\rangle|^2-|\langle u,x\rangle|^2 = \langle (vv^* -uu^*) x,x\rangle,$$
the value
$$\sup_{\|x\|\le1} |\langle v,x\rangle|^2-|\langle u,x\rangle|^2$$
coincides with the spectral norm of the rank two hermitian matrix $vv^* -uu^*$
which has zero trace and two eigenvalues $\pm\lambda$.
Since
$$2\lambda^2=\operatorname{Tr}((vv^* -uu^*)^2)=2-2|\langle u,v\rangle|^2,$$
one has $\lambda=\sqrt{1-|\langle u,v\rangle|^2}$.
A: Denote by $\alpha$, $\beta$, $\gamma$ the angles between $v$ and $x$, $u$ and $x$, $v$ and $u$ respectively (so, $\alpha,\beta,\gamma\in [0,\pi]$). Then $\alpha,\beta,\gamma$ are three planar angles of a trihedral angle (or three sides of a spherical triangle if you prefer) and they satisfy
$|\beta-\alpha|\leqslant \gamma\leqslant \alpha+\beta$ and $\alpha+\beta+\gamma\leqslant 2\pi$.
Your inequality reads as $$\cos^2\alpha-\cos^2\beta\leqslant \sin \gamma.$$
We have
$$
\cos^2\alpha-\cos^2\beta=\frac{1+\cos 2\alpha}2-\frac{1+\cos 2\beta}2=
\frac{\cos 2\alpha-\cos 2\beta}2=\sin(\beta-\alpha)\sin(\alpha+\beta)\\
\leqslant \min(\sin|\beta-\alpha|,|\sin(\alpha+\beta)|)\leqslant \sin \gamma,
$$
the last inequality follows from the above inequalities for $\alpha,\beta,\gamma$: $\gamma$ belongs to the segment between $|\beta-\alpha|$ and $\min(\alpha+\beta,2\pi-\alpha-\beta)$, and the minimal value of sine on this segment is realized on one of endpoints.
A: First reduction
After a rotation (or a selection of coordinates) we may WLOG assume that u and v are vectors in ${\bf R}^3$ and that $x=(1,0,0)^\top$. So the desired inequality is equivalent to
$$
v_1^2-u_1^2 \leq \sqrt{ 1- (u^\top v)^2 }
$$
and it suffices to prove that
$$
(u_1^2-v_1^2)^2 \leq 1 - (u^\top v)^2 
$$
which is equivalent to
$$
1 \geq (u_1^2-v_1^2)^2 + (u^\top v)^2
$$
Second reduction
Since the RHS increases if we replace each entry of $u$ or $v$ by its modulus, and this change still preserves the constraint that $u$ and $v$ are unit vectors, we may WLOG assume that all entries of these vectors are in $[0,1]$. (This is not essential, but it makes some formulas more symmetrical)
To simplify some formulas, let $a=u_1$ and $b=v_1$.
Then, by the Cauchy–Schwarz inequality (in ${\bf R}^2$) and the assumption that $u$ and $v$ are unit vectors,
$$
u_2v_2+u_3v_3 \leq \sqrt{1-a^2}\sqrt{1-b^2}
$$
and so
$$
(u^\top v)^2 =(ab+u_2v_2+u_3v_3)^2 \leq (ab+\sqrt{1-a^2}\sqrt{1-b^2})^2
$$
Hence it suffices to prove that
$$
\displaystyle 1 \geq (a^2-b^2)^2 + (ab+\sqrt{1-a^2}\sqrt{1-b})^2
$$
subject to the constraints $0\leq a\leq 1, 4\leq  b\leq 1$.
Final step
Expanding, this is equivalent to
$$
1\geq a^4 -2a^2b^2 + b^4 + a^2b^2 + 2ab\sqrt{1-a^2}\sqrt{1-b^2} + (1-a^2)(1-b^2)
$$
which rearranges to
$$
a^2(1-a^2) + b^2(1-b^2) \geq 2 ab\sqrt{1-a^2}\sqrt{1-b^2}
$$
This now follows by applying the AM-GM inequality to $a\sqrt{1-a^2}$ and $b\sqrt{1-b^2}$.
Afterthought
The use of Cauchy–Schwarz in the second stage could probably be replaced with a more geometric argument, but I don't quite have the right intuition.
