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For all $x$ and $y$ in $[0,1]^2$ $$|f(x,y)|\le\int_0^1|f_y(x,z)|\,dz\le\sqrt{\int_0^1|f_y(x,z)|^2\,dz}, $$ where $f_y(x,z):=\frac{\partial f(x,y)}{\partial y}|_{y=z}$. So, $$1=\iint_{[0,1]^2}dx\,dy\,|f(x,y)|^2\le\iiint_{[0,1]^3}dx\,dy\,dz\,|f_y(x,z)|^2 \le\|f-1\|_{H^1}^2. $$ So, any $c\in(0,1)$ will do.