lf a ,b, c,d, and p are different real numbers such that $\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right){\mathrm{p}}^2-2(\mathrm{ab}+\mathrm{bc}+\mathrm{cd})$p$+\left({\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2\right)\le 0$, then a, b, c, d are in

$\begin{array}{r}\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right){\mathrm{p}}^2-2(\mathrm{ab}+\mathrm{bc}+\mathrm{cd})\mathrm{p}+\left({\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2\right)\le 0\dots \left(\mathrm{i}\right) \\ \end{array}$  

$\text{ LHS }=\left({\mathrm{a}}^2{\mathrm{p}}^2-2\mathrm{abp}+{\mathrm{b}}^2\right)+\left({\mathrm{b}}^2{\mathrm{p}}^2-2\mathrm{bcp}+{\mathrm{c}}^2\right)$

$+\left({\mathrm{c}}^2{\mathrm{p}}^2-2\mathrm{cdp}+{\mathrm{d}}^2\right)$

$=(\mathrm{ap}-\mathrm{b}{)}^2+(\mathrm{bp}-\mathrm{c}{)}^2+(\mathrm{cp}-\mathrm{d}{)}^2\ge 0 ......(\mathrm{i})$

Since,  the  sum of squares  of real  number  is non-negative.

From Eqs.(i) and  (ii). we  get

$\begin{array}{l}(\mathrm{ap}-\mathrm{b}{)}^2+(\mathrm{bp}-\mathrm{c}{)}^2+(\mathrm{cp}-\mathrm{d}{)}^2=0\\ \Rightarrow \mathrm{ap}-\mathrm{b}=0=\mathrm{bp}-\mathrm{c}=\mathrm{cp}-\mathrm{d}\\ \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{p}\end{array}$