If $a,b,c,d$ and $p$ are distinct real number such that $\left( a^{2} + b^{2} + c^{2} \right)p^{2} - 2(ab + bc + cd)p + (b^{2} + c^{2} + d^{2}) \leq 0$, then $a,b,c,d$

Question

Dr. Abubacker · Accepted Answer

(b)
Here, $\left( a^{2} + b^{2} + c^{2} ight)p^{2} - 2(ab + bc + cd)p + (b^{2} + c^{2} + d^{2}) \leq 0$
$\Rightarrow \left( a^{2}p^{2} - 2abp + b^{2} ight) + \left( b^{2}p^{2} - 2bcp + c^{2} ight) + (c^{2}p^{2} - 2cdp + d^{2}) \leq 0$
$\Rightarrow (ap - b)^{2} + (bp - c)^{2} + (cp - d)^{2} \leq 0$
(Since sum of squares is never less than zero).
$\Rightarrow$ Each of the square is zero.
$	herefore(ap - b)^{2} = (bp - c)^{2} = (cp - d)^{2} = 0$
$\Rightarrow p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$
$	herefore\ a,b,c,d$ are in GP.

Q. If $a,b,c,d$ and $p$ are distinct real number such that $\left( a^{2} + b^{2} + c^{2} \right)p^{2} - 2(ab + bc + cd)p + (b^{2} + c^{2} + d^{2}) \leq 0$, then $a,b,c,d$

Explanation

Reviewed by Dr. Abubacker

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