jmc

Let $x=\sqrt{a+b+c-d}.$ Then $x^2=a+b+c-d,$ so $d=a+b+c-x^2,$ and we can write $$a^2+b^2+c^2+1=a+b+c-x^2+x.$$Then $$a^2-a+b^2-b+c^2-c+x^2-x+1=0.$$Completing the square in $a,$ $b,$ $c,$ and $x,$ we get $${\left(a-\frac{1}{2}\right)}^2+{\left(b-\frac{1}{2}\right)}^2+{\left(c-\frac{1}{2}\right)}^2+{\left(x-\frac{1}{2}\right)}^2=0.$$Hence, $a=b=c=x=\frac{1}{2},$ so $$d=a+b+c-x^2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\boxed{\frac{5}{4}}.$$