TITLE: Dot product properties
QUESTION [5 upvotes]: I want to prove or contradict the following claim:
If we take two vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ in $\mathbb{R}^{d}$ ($d$ isn't neccesarily 2, so geometric proofs aren't available) and the angle between them, which is defined by $\cos(\alpha_{\mathbf{v}_1,\mathbf{v}_2}) = \frac{\mathbf{v}_1^T\mathbf{v}_2}{\Vert \mathbf{v}_1 \Vert \Vert \mathbf{v}_2 \Vert}$ the following holds:

For any vector $\mathbf{u}$ s.t. $\text{sgn}(\mathbf{v}_1^T\mathbf{u}) = \text{sgn}(\mathbf{v}_2^T\mathbf{u}) = 1$ if we denote $\tilde{\mathbf{v}}_1 = \mathbf{v}_1+\mathbf{u}$ and $\tilde{\mathbf{v}}_2 = \mathbf{v}_2+\mathbf{u}$
we'll get $\alpha_{\tilde{\mathbf{v}}_1,\tilde{\mathbf{v}}_2}<\alpha_{\mathbf{v}_1,\mathbf{v}_2}$
For any vector $\mathbf{u}$ s.t. $\text{sgn}(\mathbf{v}_1^T\mathbf{u}) = \text{sgn}(\mathbf{v}_2^T\mathbf{u}) = -1$ if we denote $\tilde{\mathbf{v}}_1 = \mathbf{v}_1-\mathbf{u}$ and $\tilde{\mathbf{v}}_2 = \mathbf{v}_2-\mathbf{u}$
we'll get $\alpha_{\tilde{\mathbf{v}}_1,\tilde{\mathbf{v}}_2}<\alpha_{\mathbf{v}_1,\mathbf{v}_2}$

I am pretty confident the above holds, since I ran a lot of numerical simulations and it does seem to hold, i.e. I believe the claim needs to be proved and not contradicted.
I attempted to use the algebraic definition of cosine with some algebraic tricks (triangle inequality etc) and it didn't work, same with the generalized cosine inequality (for vectors).

REPLY [2 votes]: Both claims are false. Since we can obtain one claim from the other by replacing $u$ by $-u$, it suffices to disprove that the first claim.
Pick two linearly independent vectors $u$ and $v_1$ such that $v_1^Tu>0$. Let $v_2=2v_1$. Then $v_2^Tu>0$ but
$$
\alpha_{v_1,v_2}=0<\alpha_{\tilde{v}_1,\tilde{v}_2}.
$$
For a concrete counterexample, let
\begin{aligned}
u&=(1,1)^T,\\
v_1&=(1,0)^T,\\
v_2&=(2,0)^T,\\
\tilde{v_1}=u+v_1&=(2,1)^T,\\
\tilde{v_2}=u+v_2&=(3,1)^T.
\end{aligned}
Then
$$
\frac{v_1^Tv_2}{\|v_1\|\|v_2\|}=1
>\frac{7}{\sqrt{50}}=\frac{\tilde{v}_1^T\tilde{v}_2}{\|\tilde{v}_1\|\|\tilde{v}_2\|}
$$
and hence
$$
\alpha_{v_1,v_2}
=\arccos\frac{v_1^Tv_2}{\|v_1\|\|v_2\|}
<\arccos\frac{\tilde{v}_1^T\tilde{v}_2}{\|\tilde{v}_1\|\|\tilde{v}_2\|}
=\alpha_{\tilde{v}_1,\tilde{v}_2}.
$$
By perturbing $v_2$ slightly along a direction normal to itself, one can also obtain a counterexample in which $v_1$ and $v_2$ are not linearly dependent.