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

Question

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.

Answer 1

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} $

$AC+BD>AB+BC+CD$