Note that the constant function $f=c$ satisfies your conditions for any real $c\le-1$. So, $$\inf\int_0^1 f=-\infty.$$
Consider now the same problem but with the addition condition $f\ge0$. Then, for all $x\in[0,1]$ (with $y=1-x$), we have $$\sqrt{1-x(1-x)}\le\sqrt{f(x)f(1-x)}\le\frac{f(x)+f(1-x)}2,$$ whence $$\int_0^1 f\ge\int_0^1 dx\,\sqrt{1-x(1-x)} =\frac{1}{8} (4+\ln27)=0.9119\ldots.$$