3. Given real numbers $a, b, c, d$ satisfy $a d - b c = 1$, then the minimum value of $a^{2} + b^{2} + c^{2} + d^{2} + a c + b d$ is
3. $\sqrt{3}$

Analysis: Method 1: $a^{2}+b^{2}+c^{2}+d^{2}+a c+b d$
$$
=\left(\frac{\sqrt{3}}{2} a-\frac{b}{2}-d\right)^{2}+\left(\frac{1}{2} a+\frac{\sqrt{3}}{2} b+c\right)^{2}+\sqrt{3}(a d-b c) \geq \sqrt{3}
$$

Equality holds when $a=-\frac{c}{2}, b=-\frac{\sqrt{3}}{2} c, c^{2}=\frac{2 \sqrt{3}}{3}, d=0$.
Method 2: By the Fibonacci identity, $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=(a d-b c)^{2}+(a c+b d)^{2}$, let $x=a c+b d, \therefore y=a^{2}+b^{2}+c^{2}+d^{2}+a c+b d \geq 2 \sqrt{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}+a c+b d$ $=2 \sqrt{x^{2}+1}+x$, clearly $y>0,(y-x)^{2}=4\left(x^{2}+1\right), 3 x^{2}+2 y x+4-y^{2}=0$, discriminant $\Delta=4 y^{2}-12\left(4-y^{2}\right) \geq 0$, thus $y \geq \sqrt{3}$.