6. Let $a, b, c, d$ be real numbers, satisfying
$$
a+2 b+3 c+4 d=\sqrt{10} \text {. }
$$

Then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}$ is $\qquad$
6. 1 .

From the given equation, we have
$$
\begin{array}{l}
(1-t) a+(2-t) b+(3-t) c+(4-t) d+ \\
t(a+b+c+d)=\sqrt{10} .
\end{array}
$$

By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[(1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2}\right] .} \\
{\left[a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\right] \geqslant 10} \\
\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2} \\
\quad \geqslant \frac{10}{5 t^{2}-20 t+30}=\frac{10}{5(t-2)^{2}+10}=1 .
\end{array}
$$

Equality holds if and only if
$$
t=2, a=-\frac{\sqrt{10}}{10}, b=0, c=\frac{\sqrt{10}}{10}, d=\frac{\sqrt{10}}{5}
$$

Therefore, the minimum value sought is 1.