By the cosine theorem: \[ AC^2=AB^2+BC^2-2AB.BC cos(\angle ABC) \geq AB^2+BC^2-2AB.BC (-\frac{1}{2}) = AB^2+BC^2+AB.BC \] It's obvious $ cos(\angle ABC) \leq -\frac{1}{2}$ because $ \angle ABC \geq 120^{o} $. On the other hand: \[ AB^2+BC^2+AB.BC > AB^2 + \frac{1}{4}BC^2 + AB.BC = (AB + \frac{1}{2}BC)^2 \] So $ AC > AB + \frac{1}{2}BC $. In the same way we show that : $ BD > CD + \frac{1}{2}BC $ Adding these inequalities we obtain: $ AC + BD \geq AB+BC+CD $

Let us denote by $M$ the midpoint of $BC$. Let $K$ be a point in the ray $[AB)$ such that $BK=BM$. Since the triangle $BMK$ is isosceles we have $\angle BKM= \angle BMK$ and because $ABM$ the exterior angle of triangle $BKM$ we have that $\angle ABC=\angle ABM= \angle BMK+ \angle BKM $ so since $\angle ABC \geq 120^o$ we have that $\angle BKM+ \angle BMK \geq 120^o$. so $\angle BKM \geq 60^o$ and $ \angle BMK \geq 60^o$. Also from the given we can understand that $\angle MBK < 60^o$, so in triangle $BMK$ , $MK \leq BM=MC$. In triangle $MCK$ , $\angle MCK \leq \angle MKC$. So we have that $2 \cdot \angle MKC \geq \angle MKC+ \angle MCK= \angle BMK \geq 60^o$, so we see that $\angle MKC \geq 30^o$. $\angle BKM + \angle MKC \geq 60^o+ 30^o=90^o$, so in the triangle $AKC$ , $AC>AK=AB+\frac{BC}{2}$ . with same arguments we can show that $ BD > CD+\frac{BC}{2} $. So we are done. I solved like this in the contest and I got $0$ points put of $8$. Is my solution wrong? Do I have any mistake?

We have that $\cos \widehat {ABC} \leq -\frac 1 2$: from this and the cosine theorem we get \[|AC|=\sqrt {|AB|^2+|BC|^2-2 |AB| |BC|\cos \widehat{ABC}}\geq \sqrt {|AB|^2+|BC|^2+ |AB||BC|}\geq |AB|+\frac {|BC|} 2\] and a similar argument also shows that $|BD|\geq |CD|+\frac {|BC|} 2$; adding the two inequalities we get the result.

Let $AB=a,BC=b,CD=c,AC=p,BD=q$. WLOG $BC=1$. By LoC, $p^2 \ge a^2+b^2+ab = a^2+a+1 = (a+\frac{1}{2})^2 +\frac{3}{4}\ge(a+\frac{1}{2})^2\implies p\ge a+\frac{1}{2}$. Similary for $q$, and adding gives $p+q\ge a+b+1=a+b+c$.

ridgers wrote: In a convex quadrilateral $ABCD$ ,$\angle ABC$ and $\angle BCD$ are $\geq 120^o$. Prove that $|AC|$ + $|BD| \geq |AB|+|BC|+|CD|$. (With $|XY|$ we understand the length of the segment $XY$). See also here https://artofproblemsolving.com/community/c6h1663262p10564075