# Showing boundedness below by sobolev norm

Suppose I know that $h(f)$ is bounded by $h(f) \geq (1 -2ab)\Vert f' \Vert^2_{L^2} - \frac{4a}{b} \Vert f \Vert^2_{L^2}$ with $a,b > 0$ and $f \in H^1$. Then for any $b \leq \frac{1}{2a}$ we can find a constant $c > 0$ s.t $h(f) \geq -c \Vert f \Vert^2_{L^2}$. Let $c_\infty$ denote the optimal bound.

My question is: How do I bound $h(f) + (1+c_\infty)\Vert f \Vert^2_{L^2}$ by some $\alpha > 0$ s.t

$\alpha^2 \Vert f \Vert^2_{H^1} \leq h(f) + (1+c_\infty)\Vert f \Vert^2_{L^2}$

With $c = c_\infty$ then I can obviously bound it with the $L^2$-norm, but I dont see how its done with the Sobolev norm, how can I pull out a positive alpha from

$h(f) + (1+c_\infty)\Vert f \Vert^2_{L^2} \geq (1+c_\infty - \frac{4a}{b} \Vert f \Vert^2_{L^2})+ (1 -2ab)\Vert f' \Vert^2_{L^2}$ ?

Any help is greatly appreciated.