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 additional condition $f\ge0$. Then, for all $x\in[0,1]$ (using your condition on $f$ with $x_1=x=x_2$), we have $$f(x)\ge\sqrt{1-x^2},$$ whence $$\int_0^1 f\ge\int_0^1 dx\,\sqrt{1-x^2}=\frac\pi4=0.78539\ldots.$$
With a bit more effort, we can do better than that: for all $x\in[0,1]$ (using your condition on $f$ with $x_1=x$ and $x_2=1-x$), we have $$\sqrt{1-x(1-x)}\le\sqrt{f(x)f(1-x)}\le\frac{f(x)+f(1-x)}2,$$ whence $$\int_0^1 f\ge\int_0^1 dx\,\sqrt{1-x(1-x)} =c:=\frac{1}{8} (4+\ln27)=0.9119\ldots.$$
On the other hand, if $f(x)=c_1\sqrt{1-x(1-x)}$ for $c_1:=\sqrt{\frac{2}{\sqrt{3}}}$ and all $x\in[0,1]$, then your conditions of $f$ will hold. So, the lower bound $c$ on $\int_0^1 f$ is optimal up to the factor $c_1=1.0745\ldots.$
I would be much (and pleasantly) surprised if $\inf \int_0^1 f$ is explicitly determined (assuming additionally that $f\ge0$).