Assume that $g(re^{it}),$ and $h(re^{it})$ are smooth positive functions defined on the annulus $A=A(R,1)=\{z: R<|z|<1\}$. Assume also that $\int\_0^{2\pi}h(re^{it})dt\ge 2\pi c$ for every $r\in(R,1)$. 
Can we state that for every $r$ there is $t\_r$ so that 
$$I=\int\_A g\cdot h \, dxdy\ge 2\pi c\int\_R^1 rg(re^{it\_r})dr$$



An inequality that involves integrals
This doesn't seem true. For instance, let $f : S^1 \rightarrow \mathbb{R}$ be a bump function with total mass $2\pi$ localized near $e^{i0}=1$. Define $h(re^{it}):= f(e^{2\pi i\frac{(r-R)}{1-R}}e^{it})$. Then, for any $r$, $\int\_0^{2\pi}h(re^{it})=2\pi$, so we can set $c=1$. Now we can take $g$ to be any smooth positive function on $A$ such that $g=0$ on the support of $h$ but $\int\_R^1 rg(re^{it})dr$ never vanishes. This is certainly possible because the support of $h$ does not contain any radial line. With such choices of $g$ and $h$, LHS is 0 and RHS is always positive, so it can't be true. 


Edit : By adding an epsilon to $g$ and $h$, you can make them positive. 


