# In a convex Quadrilateral ABCD, $\angle ABC = \angle BCD = 120^{\circ}$.Prove that: $AC + BD \ge AB + BC + CD$

In a convex Quadrilateral ABCD, $\angle ABC = \angle BCD = 120^{\circ}$.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.

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