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

We have,

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

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

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

This is only possible when

$\Rightarrow (ap-b{)}^2=0\Rightarrow p=\frac{b}{a}$

Also $(bp-c{)}^2=0\Rightarrow p=\frac{c}{b}$

Similarly, $(cp-d{)}^2=0\Rightarrow p=\frac{d}{c}$

$\therefore \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$

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