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