यदि $a, b, c, d$ तथा $p$ विभिन्न वास्तविक संख्याएँ इस प्रकार हैं कि $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \leq 0$ तो दर्शाइए कि $a, b, c$ तथा $d$ गुणोत्तर श्रेणी में हैं।

Given that

$\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \,\leq \,0$       .........$(1)$

But $L.H.S.$

$=\left(a^{2} p^{2}-2 a b p+b^{2}\right)+\left(b^{2} p^{2}-2 b c p+c^{2}\right)+\left(c^{2} p^{2}-2 c d p+d^{2}\right)$

which gives $(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}\, \geq \,0$          ..........$(2)$

Since the sum of squares of real numbers is non negative, therefore, from $(1)$ and $(2),$

we have, $(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0$

or $a p-b=0, b p-c=0, c p-d=0$

This implies that $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$

Hence $a, b, c$ and $d$ are in $G.P.$