Is $b(\min(u^+,1),\min(u^+,1)) \leq b(u,u)$ equivalent to $b(u^+,u^-) \leq 0$?

Question

Let $H=H^1(\Omega)$ on a bounded smooth domain.

Suppose a symmetric coercive bilinear form $b:H \times H \to \mathbb{R}$ satisfies the condition:

If $u \in H$ then $v:= \min(u^+,1) \in H$ and $b(v,v) \leq b(u,u)$.

Is this condition equivalent to

If $u \in H$ then $b(u^+,u^-) \leq 0$

Apparently, it is true but I haven't been able to prove it.

Here $u=u^+ - u^-$ is the standard decomposition into the positive and negative parts.