Solution:
Let $O$ be the centre of the circle in which $ABCD$ is inscribed and let $R$ be its radius. Using cosine rule in triangle $AOB$, we have
$$2+\sqrt{2}=2R^{2}\left(1-\cos 135^{\circ}\right)=R^{2}(2+\sqrt{2})$$
Hence $R=1$.

Consider quadrilateral $ABCD$ as in the second figure above. Join $AC$. For $[ADC]$ to be maximum, it is clear that $D$ should be the mid-point of the arc $AC$ so that its distance from the segment $AC$ is maximum. Hence $AD = DC$ for $[ABCD]$ to be maximum. Similarly, we conclude that $BC = CD$. Thus $BC = CD = DA$ which fixes the quadrilateral $ABCD$. Therefore each of the sides $BC, CD, DA$ subtends equal angles at the centre $O$.

Let $\angle BOC = \alpha$, $\angle COD = \beta$ and $\angle DOA = \gamma$. Observe that
$$[ABCD] = [AOB] + [BOC] + [COD] + [DOA] = \frac{1}{2} \sin 135^{\circ} + \frac{1}{2}(\sin \alpha + \sin \beta + \sin \gamma)$$
Now $[ABCD]$ has maximum area if and only if $\alpha = \beta = \gamma = \left(360^{\circ} - 135^{\circ}\right)/3 = 75^{\circ}$. Thus
$$[ABCD] = \frac{1}{2} \sin 135^{\circ} + \frac{3}{2} \sin 75^{\circ} = \frac{1}{2}\left(\frac{1}{\sqrt{2}} + 3 \frac{\sqrt{3}+1}{2\sqrt{2}}\right) = \frac{5+3\sqrt{3}}{4\sqrt{2}}$$

Alternatively, we can use Jensen's inequality. Observe that $\alpha, \beta, \gamma$ are all less than $180^{\circ}$. Since $\sin x$ is concave on $(0, \pi)$, Jensen's inequality gives
$$\frac{\sin \alpha + \sin \beta + \sin \gamma}{3} \leq \sin\left(\frac{\alpha + \beta + \gamma}{3}\right) = \sin 75^{\circ}$$
Hence
$$[ABCD] \leq \frac{1}{2\sqrt{2}} + \frac{3}{2} \sin 75^{\circ} = \frac{5+3\sqrt{3}}{4\sqrt{2}}$$
with equality if and only if $\alpha = \beta = \gamma = 75^{\circ}$.