If $a,b,c$ and $d$ are four real numbers of the same sign, then the value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ lies in the interval 

Given:  $a,b,c$ and $d$ are four real numbers of the same sign.

To find: value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ lies in which interval.

The arithmetic mean is greater than geometric mean for three positive numbers $x,y,z$. This can be written as $\frac{x+y+z}{3}\ge \left(xyz\right)\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Where equality holds when $x=y=z$.

Applying $AM\ge GM$ for $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}$ and $\frac{d}{a}$,

         $$\begin{gathered} \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}}{4}\ge {\left(\frac{a}{b}\times \frac{b}{c}\times \frac{c}{d}\times \frac{d}{a}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \\ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)\ge 4 \\ \therefore \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\in \left(4,\infty \right) \end{gathered}$$

Hence, option (d) is correct.