jmc

It is clear that $0\le a\le 1,$ so $a(1-a)\ge 0.$ Then $a-a^2\ge 0,$ or $a^2\le a.$ Similarly, $b^2\le b,$ $c^2\le c,$ and $d^2\le d,$ so $$a^2+b^2+c^2+d^2\le a+b+c+d=1.$$Equality occurs when $a=1$ and $b=c=d=0,$ so the maximum value is $\boxed{1}.$