If $a, b, c, d$ and $p$ are different real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2}) \leq 0$,then show that $a, b, c$ and $d$ are in $G.P.$

Given that $(a^{2}+b^{2}+c^{2}) p^{2}-2(ab+bc+cd) p+(b^{2}+c^{2}+d^{2}) \leq 0$ $(1)$
Expanding the expression,we get:
$(a^{2}p^{2}-2abp+b^{2})+(b^{2}p^{2}-2bcp+c^{2})+(c^{2}p^{2}-2cdp+d^{2}) \leq 0$
This can be written as:
$(ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2} \leq 0$ $(2)$
Since the sum of squares of real numbers is always non-negative,the only way for the sum to be $\leq 0$ is if each term is equal to $0$:
$(ap-b)^{2} = 0, (bp-c)^{2} = 0, (cp-d)^{2} = 0$
This implies $ap=b, bp=c, cp=d$.
Thus,$\frac{b}{a} = p, \frac{c}{b} = p, \frac{d}{c} = p$.
Since the common ratio is constant,$a, b, c, d$ are in $G.P.$