If $a_1,a_2,a_3,\dots ,a_n$ are non-zero real numbers such that $\left(a_1^2+a_2^2+\dots +a_{n-1}^2\right)\left(a_2^2+a_3^2+\dots +a_n^2\right)\le {\left(a_1{a}_2+a_2{a}_3+\dots +a_{n-1}{a}_n\right)}^2$ then, $a_1,a_2,\dots ,a_n$ are in 

Using

$\left(a_1^2+a_2^2+\dots +a_n^2\right)\left(b_1^2+b_2^2+\dots +b_n^2\right)-{\left(a_1{b}_1+\dots +a_n{b}_n\right)}^2$

$=\sum _{i\ne j} {\left(a_i{b}_j-a_j{b}_i\right)}^2$,we obtain

$\begin{array}{l}\left(a_1^2+a_2^2+\dots +a_{n-1}^2\right)\left(a_2^2+a_3^2+\dots +a_n^2\right) \le {\left(a_1{a}_2+a_2{a}_3+\dots +a_{n-1}{a}_n\right)}^2\\ \Rightarrow {\left(a_1{a}_3-a_2^2\right)}^2+{\left(a_2{a}_4-a_3^2\right)}^2+\dots +{\left(a_{n-2}{a}_n-a_{n-1}^2\right)}^2\le 0\\ \Rightarrow a_1{a}_3-a_2^2=0,a_2{a}_4-a_3^2=0,\dots ,a_{n-1}{a}_n-a_{n-1}^2=0\\ \Rightarrow \frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=\dots =\frac{a_{n-1}}{a_n}\\ \Rightarrow a_1,a_2,a_3,\dots ,a_n \text{ are in G.P. }\end{array}$