Note that the constant function $f=c$ satisfies your conditions for any real $c\le-1$. So, 
$$\inf\int_0^1 f=-\infty.$$

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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 
$$\int_0^1 f\ge\int_0^1 dx\,\sqrt{1-x(1-x)} 
=\frac{1}{8} (4+\ln27)=0.9119\ldots.$$