Question

If $x_1,x_2,....x_n$ are $n$ non-zero real numbers such that$(x_1^2{+x}_2^2+x_3^2+.....x_{n+1}^2)(x_2^2+x_3^2+x_4^2+....x_n^2)\le {(x_1{x}_2+x_2{x}_3+....x_{n-1}{x}_n)}^2$then prove that $x_1,x_2,.x_n$ are in $G.P.$

Answer 1

We have $\left({)}^2\right({)}^2-({)}^2\le 0$
$(x_1^2{x}_1^2+x_1^2{x}_3^2+x_2^2{x}_2^2+)-(x_1^2{x}_1^2+{2x}_1{x}_3{x}_2^2+)\le 0$
$(x_1{x}_3-x_2^2{)}^2+(x_2{x}_4-x_3^2{)}^2+(x_3{x}_5-x_2^2{)}^2+\le 0$
Above is possible only when each term is zero.
$\therefore x_1{x}_3=x_2^2,x_2{x}_4=x_3^2,x_3{x}_5=x_4^2$
$\Rightarrow x_1,x_2,x_3,x_4,x_5,x_n$ are in $G.P.$