If a, b, c, d and p are different real number | vedclass

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(B) Given the inequality: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ ... $(i)$We can rewrite the expression on the left-hand

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$G.P.$

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(B) Given the inequality: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ ... $(i)$We can rewrite the expression on the left-hand side as:$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$This simplifies to:$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$ ... $(ii)$Since the sum of squares of real numbers is always non-negative,the only way for the sum to be less than or equal to $0$ is if each term is exactly $0$:$(ap - b)^2 = 0 \Rightarrow ap = b \Rightarrow p = b/a$$(bp - c)^2 = 0 \Rightarrow bp = c \Rightarrow p = c/b$$(cp - d)^2 = 0 \Rightarrow cp = d \Rightarrow p = d/c$Thus,$b/a = c/b = d/c = p$. This implies that the sequence $a, b, c, d$ has a constant common ratio $p$,which means $a, b, c, d$ are in $G.P.$

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

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