$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+d^{2}+(1-a)^{2}+(a-b)^{2}+ \\
(b-c)^{2}+(c-d)^{2}+(d-1)^{2}
\end{array}
$$

When the value of the above algebraic expression is minimized, what is the value of $a+b+c+d$? Prove your conclusion.
1. Arrange in descending powers of $a$,
$$
\text { original expression }=3 a^{2}-(2+2 b) a+\cdots \text {. }
$$

Therefore, when the value of the original expression is minimized,
$$
a=\frac{2+2 b}{2 \times 3}=\frac{1+b}{3} \text {. }
$$

Similarly, when the value of the original expression is minimized, we have
$$
d=\frac{1+c}{3}, b=\frac{a+c}{3}, c=\frac{d+b}{3} \text {. }
$$

Solving these equations simultaneously gives $b=c=\frac{1}{5}, a=d=\frac{2}{5}$.
Therefore, when the value of the original expression is minimized, $a+b+c+d$ equals $\frac{2}{5}+\frac{1}{5}+\frac{1}{5}+\frac{2}{5}=\frac{6}{5}$.