Question

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

Answer 1

$(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0,$

$\Rightarrow a^2{p}^2+b^2{p}^2+c^2{p}^2-2abp-2bcp-2cdp+b^2+c^2+d^2\le 0$

$\Rightarrow (ap{)}^2+b^2-2abp+(bp{)}^2+c^2-2bcp+(cp{)}^2+d^2-2cdp\le 0$

$\Rightarrow \left[\right(ap{)}^2+(b{)}^2-2\times ap\times b]+\left[\right(bp{)}^2+(c{)}^2-2\times bp\times c]+\left[\right(cp{)}^2+(d{)}^2-2\times cp\times d]$

$\Rightarrow (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\le 0$

Square of a real number is zero or always positive So, sum of square of real number
is zero or always positive

$\therefore (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2=0$

$(ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2=0$

$\text{Thus},(ap-b)=0,(bp-c)=0\&(cp-d)=0$

$(ap-b)=0\Rightarrow ap=b$

$\Rightarrow \frac{b}{a}=p...\left(i\right)$

$(bp-c)=0\Rightarrow bp=c$

$\Rightarrow \frac{c}{b}=p...\left(ii\right)$

$(cp-d)0\Rightarrow cp=d$

$\Rightarrow \frac{d}{c}=p$ $..\left(iii\right)$

From $\left(i\right),\left(ii\right)\&\left(iii\right)$

$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$ (common ratio)

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