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? 



Controlling angles between vectors using sum of subvector angles?
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. 


