If Dot product of $\mathbf{a}(t)$ and derivative of $\mathbf{b}(t)$ is negative implies dot product itself is negative with some conditions.

Question

Given two vectors $\mathbf{a}(t) = (r(t), s(t), -v(t), -u(t))$ and $\mathbf{b}(t) = (p(t), q(t), w(t), z(t))$, along with accompanying conditions:

1. $r(t) + s(t) - v(t) - u(t) \leq 0$
2. $r(t)p'(t) + s(t)q'(t) -v(t)w'(t) - u(t)z'(t) \leq 0$ (Dot product of $\mathbf{a}(t)$ and derivative of $\mathbf{b}(t)$)
3. $p(t) + z(t) \leq q(t) + w(t)$ or $p(t) + z(t) - q(t) - w(t) \leq 0$

My intuition says the dot product of $\mathbf{a}(t)$ and $\mathbf{b}(t)$ will be negative. I could not prove it anyway.

Is there a more general formulation or further conditions to consider before this statement can be held?

Note: Here each p(t), q(t), w(t), z(t), r(t), s(t), v(t) and u(t) are positive real valued functions of t.