Ptolemy s inequality and triangle inequality and by enlarging Angle BAC , without changing length AB and AC the inequality got stronger.

Invert through $A$ with radius $r$, such that $X$ is sent to $X'$. Then the required inequality to prove is equivalent to \[ C'D'+B'C' < AD'+AB'. \]Now, since $ABCD$ is convex and $C$ lies outside the circumcircle of $\triangle ABD$, the point $C'$ must lie inside $\triangle AB'D'$, and the result follows.

Let $C'$ be the intersection of $(ABD)$ and $AC$. Thus, $\angle BC'D=180^\circ-\angle DAB>\angle BCD$. If $C$ lies inside $(ABC'D)$, then $\angle BCD$ is sum of two arcs, where one of them is $BAD$, this would imply that $\angle BCD>\angle BC'D$, hence $C$ lies outside $(ABC'D)$, therefore it lies on the ray $AC'$ beyond $C'$. Note that $$AB\cdot DC'+AB\cdot CC'>AB\cdot CD$$and $$AD\cdot BC'+AD\cdot CC'>AD\cdot BC$$by triangle inequality. By Ptolemy's theorem, we have $$AB\cdot C'D+AD\cdot BC'=AC'\cdot BD.$$Thus, \begin{align*}AC'\cdot BD+(AB+AD)\cdot CC' &=AB\cdot C'D+AD\cdot BC'+(AB+AD)\cdot CC' \\&>AB\cdot CD+AD\cdot BC.\end{align*}On the other hand, by triangle inequality, $$AC(AB+AD)=(AC'+CC')(AB+BD)>AC'\cdot BD+(AB+AD)\cdot CC',$$we are done. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,c,D,C; O=(0,0);B=dir(110);A=dir(200);D=dir(300);c=dir(360);path w=circumcircle(A,B,c);C=(1.2c-0.2A); draw(A--B--c--D--cycle,deep);draw(w,deep);draw(B--D,deep);draw(A--C,deep);draw(C--B,deep);draw(C--D,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C'$",c,dir(c)); dot("$D$",D,dir(D)); dot("$C$",C,dir(C)); [/asy][/asy]