Let \[\angle ABD=x; \angle ACD=y; \angle BAC=x'; \angle BDC=y';\] and \[AB=a; BC=b; CD=c; DA=d, AC=m; BD=n.\] Reflect $B$ about $AD$ to get $B'$. So, triangle $BAD\cong B'AD$. \[\Longrightarrow B'A=a; B'D=n;\angle AB'D=x.\] Also, \[S_{ABD}+S_{ACD}=S_{AB'D}+S_{ACD}=S_{AB'DC}.\] Now, reflect $A$ about $BC$ to get $A'$. By similar reasonings, \[S_{BAC}+S_{BDC}=S_{BA'CD}\] So, we are required to prove that \[S_{AB'DC}>S_{BA'CD}\] Note that quadrilaterals $AB'DC$ and $BA'CD$ have sides $a, m, c, n$ each and opposite angles $x, y$ and $x', y'$ respectively. Now (I am not sure about this step, but I hope it is right), \[x+y>x'+y'\Longrightarrow \cos^2(x+y)<\cos^2(x'+y')\Longrightarrow -\cos^2(x+y)>-\cos^2(x'+y')\] \[\Longrightarrow -amcn\cos^2(x+y)>-amcn\cos^2(x'+y')\] If \[s=\frac{a+m+c+n}{2}\text{ and }T=(s-a)(s-m)(s-c)(s-n),\text{ we have }\] \[T-amcn\cos^2(x+y)>T-amcn\cos^2(x'+y')\] By Brahmagupta's formula, \[S_{AB'DC}=T-amcn\cos^2(x+y); S_{BA'CD}=T-amcn\cos^2(x'+y')\] and so we have \[S_{AB'DC}>S_{BA'CD}\] as we are required to prove.