Moving $A$ along the semicircle away from $B$ and $D$ along the semicircle away from $C$ until $AD$ is a diameter can only increase $a^2 + b^2 + c^2 + d^2 + abc + bcd$, so we may suppose that $AD$ is a diameter of the semicircle. Let $e = CE = \sqrt{c^2 + d^2 - 2cd \cos D}$. We first claim that $a^2 + b^2 + e^2 + abe = 4$. $\angle ABE = \angle ACE = \frac{\pi}{2}$, so $BE = \sqrt{4 - a^2}$ and $AC = \sqrt{4 - e^2}$. By Ptolemy's theorem, $2b + ae = \sqrt{(4 - a^2)(4 - e^2)}$, so $4b^2 + 4abe + a^2 e^2 = 16 - 4a^2 - 4e^2 + a^2 e^2$, which rearranges to $a^2 + b^2 + e^2 + abe = 4$. It is therefore sufficient to show that \[ a^2 + b^2 + e^2 + abe > a^2 + b^2 + c^2 + d^2 + abc + bcd, \] or \[ a^2 + b^2 + c^2 + d^2 - 2cd \cos D + ab \sqrt{c^2 + d^2 - 2cd \cos D} > a^2 + b^2 + c^2 + d^2 + abc + bcd, \] or \begin{align} -2cd \cos D + ab \sqrt{c^2 + d^2 - 2cd \cos D} > abc + bcd. \end{align} We note that $\angle D \geq \frac{\pi}{2}$ (since the angle is contained in a semicircle), so $\sqrt{c^2 + d^2 - 2cd \cos D} > c$, whence \begin{align*} ab \sqrt{c^2 + d^2 - 2cd \cos D} > abc. \tag{2} \end{align*} Furthermore, \[ \angle D \&= \pi - \angle CAE = \pi - (\frac{\pi}{2} - \angle CEA) = \frac{\pi}{2} + \angle CEA = \frac{\pi}{2} + (\pi - \angle ABC) = \frac{3\pi}{2} - \angle B, \] so $-\cos D = \sin B$. By the law of sines, $\frac{AC}{\sin B} = 2$, so $AC = 2 \sin B = -2 \cos B$. Because $\angle ABC > \pi$, $AC > BC = b$, whence \begin{align*} -2cd \cos B > bcd \tag{3}. \end{align*} Adding (2) and (3) yields (1), so we are done.

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Another solution: Again, we suppose that $AD$ is a diameter of the semicircle. Let $e = CE$ and $f = CA$. From a result of the previous post, $a^2 + b^2 + e^2 + abe = 4$ and $f^2 + c^2 + d^2 + fcd = 4$. Adding these gives $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + abe + fcd = 8$. Again, since $\angle B, D > \pi$, we have $e > c$ and $f > b$. In addition, $e^2 + f^2 = 4$ by the Pythagorean theorem. Hence, $4 = 8 - (e^2 + f^2) = a^2 + b^2 + c^2 + d^2 + abe + fcd > a^2 + b^2 + c^2 + d^2 + abc + bcd$.