How to show this weak formulation has unique solution?

Suppose $a$ is a bounded and coercive bilinear form on a Hilbert space $H$ and that $b$ is a bounded bilinear form on $H$ and $\ell$ is a bounded linear function also on $H$.

How do I show that: For sufficiently small $\varepsilon > 0$, the equation $$a(u^\epsilon, v) + \varepsilon b(u^\varepsilon, v) = \ell(v), \quad \mathit{ for\,\,all}\,\, v \in H,$$ has a unique solution $u^\varepsilon$.

I can't apply Lax-Milgram because $b$ is not necessarily coercive. What's the general technique for these kinds of problems?

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If $\epsilon$ is small enough, $a+\epsilon b$ will be coercive.