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This is a technical question coming out of my research.

Let $\angle(\cdot, \cdot)$ be the angle ($\in [0, \pi]$) between vectors. Consider two vectors $u, v$ in $\mathbb R^3$. Is it true that $$ \angle(u, v) \le \sum_{\Omega} \angle(u_{\Omega}, v_{\Omega})? $$ Here the $\Omega$ indexes length-two subvectors. My research problem only requires the version when $u, v \in \mathbb R_{++}^3$ (i.e., the positive orthant), but I suspect the general version might hold.

Also, does this generalize to $\mathbb R^n$, i.e., $\mathbb R^3$ is changed to $\mathbb R^n$, but we still only use length-two vectors on the right?

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  • $\begingroup$ What is $\mathbb{R}_{++}$? $\endgroup$
    – Amir Sagiv
    Commented Jul 16, 2018 at 20:01
  • $\begingroup$ @AmirSagiv The positive orthant. Edited, thanks! $\endgroup$
    – Ju Sun
    Commented Jul 16, 2018 at 20:10
  • $\begingroup$ you mean that we choose a coordinate 2-plane $\Omega$ by all 3 possible ways and $u_{\Omega}$ is a projection of $u$ onto $\Omega$? $\endgroup$
    – Fedor Petrov
    Commented Jul 16, 2018 at 20:22
  • $\begingroup$ @FedorPetrov Exactly! $\endgroup$
    – Ju Sun
    Commented Jul 16, 2018 at 20:23

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In the positive orthant you may use the following argument.

We have $\sin \angle(u,v)=\frac{\|u\times v\|}{\|u\|\cdot \|v\|}$, where $\times$ denotes the vector product. Note that the coordinates of $u\times v$ are exactly $\pm \|u_{\Omega}\times v_{\Omega}\|$. Combined with Pythagoras and obvious inequalities like $\|u_{\Omega}\|\leqslant \|u\|$ we get $\sin^2 \angle(u,v)\leqslant \sum_{\Omega} \sin^2 \angle(u_{\Omega},v_{\Omega})$. Assume that however $\sum \angle(u_{\Omega},v_{\Omega})<\angle (u,v)$. Note that whenever $0\leqslant x,y, x+y\leqslant \pi/2$ we have $\sin^2(x+y)=(\sin x\cos y+\cos x\sin y)^2= \sin^2 x+\sin^2 y+2\sin x\sin y \cos(x+y)\geqslant \sin^2 x+\sin^2 y.$

Applying this twice we get $\sin^2 \sum \angle(u_{\Omega},v_{\Omega})\geqslant \sum \sin^2 \angle(u_{\Omega},v_{\Omega})\geqslant \sin^2 \angle(u,v)$, a contradiction.

ED by Ju: The result can be generalized to $\mathbb R^n_{++}$ using the Lagrange's identity, as discussed in the comments below.

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  • $\begingroup$ Very nice! You would make your proof even shorter by avoiding "contradiction" in the reasonning. $\endgroup$
    – Luc Guyot
    Commented Jul 16, 2018 at 21:36
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    $\begingroup$ I need the sum to be less than $\pi/2$, without contradiction this is not always so $\endgroup$
    – Fedor Petrov
    Commented Jul 16, 2018 at 21:37
  • $\begingroup$ @FedorPetrov Many thanks, Fedor! That solves the $\mathbb R^3$ case. $\endgroup$
    – Ju Sun
    Commented Jul 16, 2018 at 21:55
  • $\begingroup$ @FedorPetrov Also I realize the argument can be generalized to $\mathbb R^n$ using Lagrange's identity. This also avoids the difficulty of directly generalizing the cross product. $\endgroup$
    – Ju Sun
    Commented Jul 16, 2018 at 22:16
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    $\begingroup$ @JuSun, If you know how to generalize it, I suggest you add this as an answer to your own question. This way everyone can benefit from it! $\endgroup$
    – Amir Sagiv
    Commented Jul 17, 2018 at 6:19

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