In a convex Quadrilateral ABCD, $\measuredangle ABC = \measuredangle BCD = 120^{\circ}$.Prove that: $ AC + BD \ge AB + BC + CD$ In a convex Quadrilateral ABCD, $\measuredangle ABC  = \measuredangle BCD = 120^{\circ}$.Prove that:  
$$ AC + BD \ge AB + BC + CD$$
My attempt
Tried to use cosine formula twice' i.e ($\triangle ABC$ and $\triangle BCD$) and tried to make and inequality.But couldn't prove.

I suppose the question is :
$AC+BD > AB+BC+CD$
$AC=\sqrt{AB^2+BC^2+AB*BC}=\sqrt{(AB+\dfrac{BC}{2})^2+\dfrac{3BC^2}{4}}>AB+\dfrac{BC}{2} $
$BD=\sqrt{CD^2+BC^2+CD*BC}=\sqrt{(CD+\dfrac{BC}{2})^2+\dfrac{3BC^2}{4}}>CD+\dfrac{BC}{2} $
$AC+BD>AB+BC+CD$
