jmc

Let $$f(a,b,c,d)=a^2+b^2+c^2+d^2-(ab+\lambda bc+cd).$$For fixed values of $b,$ $c,$ and $d,$ $f(a,b,c,d)$ is minimized when $a=\frac{b}{2}.$ Similarly, for fixed values of $a,$ $b,$ $c,$ $f(a,b,c,d)$ is minimized when $d=\frac{c}{2}.$ Thus, it suffices to look at the case where $a=\frac{b}{2}$ and $d=\frac{c}{2},$ in which case the given inequality becomes $$\frac{5b^2}{4}+\frac{5c^2}{4}\ge \frac{b^2}{2}+\lambda bc+\frac{c^2}{2},$$or $5b^2+5c^2\ge 2b^2+4\lambda bc+2c^2.$ This reduces to $$3b^2+3c^2\ge 4\lambda bc.$$Taking $b=c=1,$ we find $6\ge 4\lambda ,$ so $\lambda \le \frac{3}{2}.$

On the other hand, if $\lambda =\frac{3}{2},$ then the inequality above becomes $$3b^2+3c^2\ge 6bc,$$which holds due to AM-GM. Therefore, the largest such $\lambda$ is $\boxed{\frac{3}{2}}.$