Consider functions in the Sobolev space $W^{1,2}_0 (I)$ where $I \subset \mathbb{R}$ is an interval. Define a bilinear form $$ A: W^{1,2}_0 (I)\times W^{1,2}_0 (I) \rightarrow \mathbb{R}$$ as $$ A(f, g) := \int_{I} f' g' h dx ,$$ where the "weight" $h$ is positive and bounded (almost everywhere wrt the Lebesgue measure) on $I$.
I am struggling to show that $A$ is coercive, that is, for every $f \in W^{1,2}_0 (I)$, $$ A(f, f) \ge \alpha (\Vert f \Vert_{L^2 (I)}^2 + \Vert f' \Vert_{L^2(I)}^2)$$ for some constant $\alpha > 0$. In particular, I am not sure how to get rid of the function $h$. If $h$ was bounded from below (almost everywhere) by some positive constant, then this would be easy. But what about if it is bounded from below (almost everywhere) only by zero?
Then, it is simply not true. Pick $h(x)= x$ on the interval $[0,1]$ which satisfies your assumptions. Now chose $f_n(x)=(1-nx)\mathbf{1}_{[0,\frac1n]}$. $A(f_n,f_n) =\frac12$, while $\|f'_n\|_{L^2}^2 \sim n$.