Label the lengths $a$, $b$, $c$, $d$, $e$, and $f$ in the order they were mentioned. It suffices to prove $$(a+b)^2<(c+d)^2+(e+f)^2$$However, after applying Ptolemy's inequality this reduces to showing that $$a^2+b^2<c^2+d^2+e^2+f^2$$Let $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$, and $D=(d_1,d_2)$. Then it suffices to split this up and show that $$(a_1-b_1)^2+(c_1-d_1)^2<(a_1-c_1)^2+(b_1-d_1)^2+(a_1-d_1)^2+(b_1-c_1)^2\iff 0<(a_1+b_1-c_1-d_1)^2$$However we have a slight problem when $a_1+b_1=c_1+d_1$ and $a_2+b_2=c_2+d_2$ as it would force the inequality to be not strict, but this would make $ABCD$ a rectangle.