If $ABCD$ is a convex quadrilateral with $AB + BD < AC + CD$, prove that $AB < AC$. 
If $ABCD$ is a convex quadrilateral with $AB + BD < AC + CD$, prove that $AB < AC$.

If we could use the triangle inequality combined with the given, I believe we can achieve something. However, I'm not sure if I am adding inequalities correctly.
 A: 
Since $AB+BD<AC+CD$, the ellipse with foci at $A,D$ through $B$ lies inside the confocal ellipse through $C$. 
Assuming $AC\leq AB$, the point $C$ has to lie in the circle centered at $A$ with radius $AB$. Since $ABCD$ is convex, $C$ has to lie in the angle $\widehat{DAB}$, but the intersection between the angle $\widehat{DAB}$ and the circle centered at $A$ is in the interior of the ellipse through $C$, leading to a contradiction.
A: Quoting $\it Elements$, one has
${\bf Proposition~ I.18.}$ In any triangle the greater side subtends the greater angle.
${\bf Theorem.}$ In a convex quadrilateral $ABCD$, $$AB+BD<AC+CD\Rightarrow AB<AC.$$
$\it Proof.$ We prove the contrapostive statement. Assume $$AB\geq AC. \qquad(1)$$ Then $\angle BCA\geq \angle ABC$ by Prop.I.18 (the equality is the case of isosceles triangles, Prop.I.5). Since $ABCD$ is convex, $D$ lies in the region nonnegatively spanned by $\overrightarrow{BA}$ and $\overrightarrow{BC}$, exterior to triangle $ABC$. It follows that $$\angle BCD\geq\angle BCA\geq \angle ABC\geq\angle DBC.$$ Hence by Prop.I.18 in triangle DBC, one has $$BD\geq CD.\qquad (2)$$
Then by (1) and (2), $$AB+BD\geq AC+CD.$$ QED
