jmc

Setting $x=y=z=0,$ we get $$f(0)+f(0)-f(0{)}^2\ge 1,$$so $f(0{)}^2-2f(0)+1\le 0.$ Then $(f(0)-1{)}^2\le 0,$ which forces $f(0)=1.$

Setting $x=y=z=1,$ we get $$f(1)+f(1)-f(1{)}^2\ge 1,$$so $f(1{)}^2-2f(1)+1\le 0.$ Then $(f(1)-1{)}^2\le 0,$ which forces $f(1)=1.$

Setting $y=z=0,$ we get $$f(0)+f(0)-f(x)f(0)\ge 1,$$so $f(x)\le 1$ for all $x.$

Setting $y=z=1,$ we get $$f(x)+f(x)-f(x)f(1)\ge 1,$$so $f(x)\ge 1$ for all $x.$

This tells us that the only possible function is $f(x)=1.$ We readily see that this function works, so there is only $\boxed{1}$ possible function $f(x).$