Using Cauchy-Schwarz Inequality

$a+2b+3c+4d=\sqrt{10}\Longrightarrow (1-t)a+(2-t)b+(3-t)c+(4-t)d+t(a+b+c+d)=\sqrt{10}$ ,for any real number t.

By Cauchy-Schwarz Inequality,

$\displaystyle\Big( (1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2} \Big)\Big((a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\Big)$ $\ge \Big((1-t)a+(2-t)b+(3-t)c+(4-t)d+t(a+b+c+d)\Big)^{2}=10$

$\displaystyle\Longrightarrow a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\ge \frac{10}{5(t-2)^{2}+10} =f(t)$ .

Since it holds for any real number t, so the left side should larger than or equal to the maximum of f(t),which is $f(2)=1$ .

In case $t=2,a=-\frac{\sqrt{10}}{10},b=0, c=\frac{\sqrt{10}}{10},d=\frac{\sqrt{10}}{5}$ , the left side is exactly equal to 1.

So the minimum is $\boxed{1}$ .