First of all denote $AB,BC,CD,DE,AE$ BY $a,b,c,d,e$ respectively. $\angle{AED}=60$ Extend $AB$ and $CD$ to meet at $F$. Note that $\triangle{BCF}$ is equilateral. Also, $AFDE$ is a parallelogram. Thus, $AE=b+c$ , $ED=a+b$ By cosine rule, $AC^2=a^2+b^2+ab$ , $BD^2=b^2+c^2+bc$ To prove, $16(a^2+b^2+ab)(b^2+c^2+bc)\geq{9(a+b)^2(b+c)^2}$ $(1-\frac{ab}{(a+b)^2})(1-\frac{bc}{(b+c)^2})\geq{\frac{9}{16}}$ $(a+b)^2\geq{4ab}$ $(1-\frac{ab}{(a+b)^2})\geq{\frac{3}{4}}$ $(1-\frac{bc}{(b+c)^2}\geq{\frac{3}{4}}$ Thus, multiplying we get that which is required.

Very nice solution! Thank you.