set $x=\frac{a}{d}, y=\frac{d}{c}, z=\frac{c}{b}, t=\frac{b}{a}$ it is clear that $xyzt=1$ and by dividing $2|d-a|\leq |a|$ by $|d|$, we conclude that $|x|\geq 2|x-1|$. we can similarly find such an inequality for $y,z,t$. we want to prove $|x+y+z+t|\geq \frac{7}{2}$. now set $x=x_1 +ix_2$ the previous inequality turns into $x_1 ^2 +x_2 ^2\geq 4((1-x_1)^2 +x_2^2)$. rearranging gives us $8x_1^2-4\geq 3(x_1^2 +x_2 ^2)=3 |x|^2\geq0$. similarly for $y,z,t$. now by AM-GM we have : $$(\frac{8(x_1+y_1+z_1+t_1)-16}{4})^4\geq \prod{(8x_1-4)}\geq 3^4 \cdot |xyzt|^2=3^4 \rightarrow x_1+y_1+z_1+t_1\geq \frac{7}{2}$$finally, note that $|x+y+z+t|\geq Re(x+y+z+t)= x_1+y_1+z_1+t_1\geq \frac{7}{2}$ we are done

matinyousefi wrote: If $a, b, c$ and $d$ are complex non-zero numbers such that $$2|a-b|\leq |b|, 2|b-c|\leq |c|, 2|c-d| \leq |d| , 2|d-a|\leq |a|.$$Prove that $$\frac{7}{2} <\left| \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d} \right| .$$ Proposed by Navid Safaei