If a, b, c, d and p are different real numbers such that:


If abcd and p are different real numbers such that:

(a2 + b2 + c2p2 − 2 (ab + bc + cdp + (b2 + c2 + d2) ≤ 0, then show that abc and d are in G.P.


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

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

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

$\Rightarrow(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2} \leq 0$

$\Rightarrow(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0$

$\Rightarrow(a p-b)^{2}=0$

$\Rightarrow p=\frac{b}{a}$

Also, $(b p-c)^{2}=0$

$\Rightarrow p=\frac{c}{b}$

Similiarly, $\Rightarrow(c p-d)^{2}=0$

$\Rightarrow p=\frac{d}{c}$

$\therefore \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$

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

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