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+a^2) 0, then show that a, b, c and d are in G.P.

(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+a^2) 0, (a^2p^2+b^2p^2+c^2p^2)-2(abp+bcp+cdp)+(b^2+c^2+d^2) 0 (a^2p^2-2abp+b^2)+(b^2p^2-2bcp+c^2)+(c^2p^2-2cdp+d^2) 0 (ap-b)^2+(bp-c)^2+(cp-d)^2 0 (ap-b)^2+(bp-c)^2+(cp-d)^2=0 (ap-b)^2=0 = b a Also, (bp-c)^2=0 = c b Similiarly, (cp-d)^2=0 = d c b a = c b = d c Thus, a, b, c and d are in G.P.

If a, b, c, d and p are different real numbers such that: (a^2+b^…

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If a, b, c, d and p are different real numbers such that:
$(\text{a}^2+\text{b}^2+\text{c}^2)\text{p}^2-2(\text{ab}+\text{bc}+\text{cd})\text{p}+(\text{b}^2+\text{c}^2+\text{a}^2)\le0,$ then show that a, b, c and d are in G.P.

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Answer

$(\text{a}^2+\text{b}^2+\text{c}^2)\text{p}^2-2(\text{ab}+\text{bc}+\text{cd})\text{p}+(\text{b}^2+\text{c}^2+\text{a}^2)\le0,$
$\Rightarrow(\text{a}^2\text{p}^2+\text{b}^2\text{p}^2+\text{c}^2\text{p}^2)-2(\text{a}\text{b}\text{p}+\text{b}\text{c}\text{p}+\text{c}\text{d}\text{p})+(\text{b}^2+\text{c}^2+\text{d}^2)\le0$
$\Rightarrow(\text{a}^2\text{p}^2-2\text{a}\text{b}\text{p}+\text{b}^2)+(\text{b}^2\text{p}^2-2\text{b}\text{c}\text{p}+\text{c}^2)+(\text{c}^2\text{p}^2-2\text{c}\text{d}\text{p}+\text{d}^2)\le0$
$\Rightarrow(\text{a}\text{p}-\text{b})^2+(\text{b}\text{p}-\text{c})^2+(\text{c}\text{p}-\text{d})^2\le0$
$\Rightarrow(\text{a}\text{p}-\text{b})^2+(\text{b}\text{p}-\text{c})^2+(\text{c}\text{p}-\text{d})^2=0$
$\Rightarrow(\text{a}\text{p}-\text{b})^2=0$
$\Rightarrow\text{p}=\frac{\text{b}}{\text{a}}$
Also, $(\text{bp}-\text{c})^2=0$
$\Rightarrow\text{p}=\frac{\text{c}}{\text{b}}$
Similiarly, $\Rightarrow(\text{cp}-\text{d})^2=0$
$\Rightarrow\text{p}=\frac{\text{d}}{\text{c}}$
$\therefore\frac{\text{b}}{\text{a}}=\frac{\text{c}}{\text{b}}=\frac{\text{d}}{\text{c}}$
Thus, a, b, c and d are in G.P.

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