jmc

Note that $$ab+bc+cd\le ab+bc+cd+da=(a+c)(b+d).$$By AM-GM, $$(a+c)(b+d)\le {\left(\frac{(a+c)+(b+d)}{2}\right)}^2=\frac{1}{4}.$$Equality occurs when $a=0,$ $b=\frac{1}{2},$ $c=\frac{1}{2},$ and $d=0,$ so the maximum value of $ab+bc+cd$ is $\boxed{\frac{1}{4}}.$