If a, b, c, d and p are distinct real numbers | vedclass

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(C) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \leq 0$. This can be rewritten as $(a^2p^2 - 2abp + b^2) + (b

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$a, b, c, d$ are in geometric progression

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(C) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \leq 0$. This can be rewritten as $(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \leq 0$. This simplifies to $(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \leq 0$. Since the sum of squares of real numbers is non-negative,the only way for the sum to be $\leq 0$ is if each term is exactly $0$. Thus,$ap - b = 0$,$bp - c = 0$,and $cp - d = 0$. This implies $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$. This shows that $a, b, c, d$ are in geometric progression.

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \leq 0$,then:

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