I am trying to prove that the dot product of two timelike vectors cannot be $0$.
The timelike vectors defined as,
$$g(\vec{v_1}, \vec{v_1}) = \vec{v_1} \cdot \vec{v_1} <0$$ $$g(\vec{v_2}, \vec{v_2}) = \vec{v_2} \cdot \vec{v_2} <0$$
And the ortogonality is,
$$g(\vec{v_1}, \vec{v_2}) = \vec{v_1} \cdot \vec{v_2} = 0$$
where $g(\vec{e}_{\mu}, \vec{e}_{\nu})=g_{\mu \nu} = \eta_{\mu \nu} = diag(-1,1,1,1)$
So I tried something like,
$\vec{v_1} \cdot \vec{v_1} = -(v_1^0)^2 + (v_1^1)^2 + (v_1^2)^2 + (v_1^3)^2 < 0$ So $$(v_1^0)^2 > (v_1^1)^2 + (v_1^2)^2 + (v_1^3)^2 = V_1^2 ~~(1)$$ Similarly $$(v_2^0)^2 > (v_2^1)^2 + (v_2^2)^2 + (v_2^3)^2 = V_2^2 ~~(2)$$ And
$\vec{v_1} \cdot \vec{v_2} = -(v_1^0v_2^0) + (v_1^1v_2^1) + (v_1^2v_2^2)+ (v_1^3v_2^3)$
and $L =(v_1^1v_2^1) + (v_1^2v_2^2)+ (v_1^3v_2^3)$
Then I multiplied (1) and (2) to get
$$(v_1^0)^2(v_2^0)^2 > V_1^2 V_2^2$$ I am not sure but then I said,
$$(v_1^0)(v_2^0) > V_1 V_2 > L$$ so $\vec{v_1} \cdot \vec{v_2} <0$.
Is this can be considered as a valid proof. Or is there a more elegant one
You started out okay. $$\begin{align} (v_1^0)^2 &> (v_1^1)^2 + (v_1^2)^2 + (v_1^3)^2=||\vec{w_1}||^2\tag1\\ (v_2^0)^2 &> (v_2^1)^2 + (v_2^2)^2 + (v_2^3)^2=||\vec{w_2}||^2\tag2 \end{align}$$ where $\vec{w_i} = v_i^1\mathbf{i}+v_i^2\mathbf{j}+v_i^3\mathbf{k},\ i=1,2$ are vectors in $\mathbb{R}^3.$
Suppose by way of contradiction that $v_i\cdot v_2=0$, that is $$v_1^0v_2^0=v_1^1v_2^1+v_1^2v_2^2+v_1^3v_2^3=\vec{w_1}\cdot \vec{w_2}\tag3$$
Multiplying $(1)$ by $(2)$ gives $$(v_1^0)^2(v_2^0)^2>||\vec{w_1}||^2||\vec{w_2}||^2\tag4$$
Squaring both sides of $(3)$ gives $$(v_1^0)^2(v_2^0)^2=(\vec{w_1}\cdot\vec{w_2})^2\tag5$$
Now, $(4)$ and $(5)$ together contradict the Cauchy-Schwarz inequality.