According to Cauchy-Schwarz inequality
$$\left( \frac{(a+c)^2}{ad+bc} + \frac{(b+d)^2}{ac+bd} + 4 \right) \left( (ad+bc) + (ac+bd) + 1 \right) \geq (a+c+b+d+2)^2 = 16 \implies \frac{(a+c)^2}{ad+bc} + \frac{(b+d)^2}{ac+bd} + 4 \geq \frac{16}{ad+bc+ac+bd+1}.$$
Set $a+b=x$, $c+d=y$ (therefore $x+y=2$, $x^2+y^2=4-2xy$). It suffices to show that
$$\frac{16}{ad + bc + ac + bd + 1} = \frac{16}{xy + 1} \geq 4 \left( \frac{x+1}{y+1} + \frac{y+1}{x+1} \right).$$
This is equivalent to
$$\begin{align*} 4(x+1)(y+1) &\geq (xy+1)((x+1)^2 + (y+1)^2) \\ \Leftrightarrow \quad 4(xy+3) &\geq (xy+1)(10-2xy) \\ \Leftrightarrow \quad 4xy+12 &\geq 8xy-2(xy)^2+10 \\ \Leftrightarrow \quad (xy)^2+1 &\geq 2xy \\ \Leftrightarrow \quad (xy-1)^2 &\geq 0. \end{align*}$$
Which is true, hence the claim of the problem. ■