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

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Solution

The given relation can be written as
$(a^{2} p^{2} - 2 a b p + b^{2}) + () + () \leq 0$
or $(a p - b)^{2} + (b p - c)^{2} + (c p - d)^{2} \leq 0$
Since $a, b, c, d$ and $p$ are all real, the inequality $(1)$ is possible only when each of the factors is zero i.e. $a p - b = 0$ , $b p - c = 0$ , $c p - d = 0$
or $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$ or $a, b, c, d$ are in $G . P .$

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