Questions & Answers

If ${x_1},{x_2},....{x_n}$ are $n$ non-zero real numbers such that
$({x_1}^2 + {x_2}^2 + {x_3}^2 + .......{x^2}_{n + 1})({x_2}^2 + {x_3}^2 + .....{x_n}^2) \leqslant {({x_1}{x_2} + {x_2}{x_3} + ...{x_{n - 1}}{x_n})^2}$ then prove that${x_1},{x_2},..{x_n}$ are in GP

Answer Verified Verified
Hint: Apply required Geometric progression rules and formulas required to solve the non zero real numbers.

Let us consider
$({x_1}^2 + {x_2}^2 + {x_3}^2 + .....{x_{n + 1}}^2) \to (a)$
$({x^2}_2 + {x^2}_3 + .....{x^2}_n) \to (b)$
$({x_1}{x_2} + {x_2}{x_3} + ......{x_{n - 1}}{x_n}) \to (c)$
If we observe the given expression it is in the form${(a)^2}{(b)^2} - {(c)^2} \leqslant 0$
From this we can write
  ({x^2}_1{x^2}_1 + {x^2}_1{x^2}_3 + {x^2}_2{x^2}_2 + ..) - ({x^2}_1{x^2}_1 + 2{x_1}{x_3}{x^2}_2) \leqslant 0 \\
  {({x_1}{x_3} - {x^2}_2)^2} + {({x_2}{x_4} - {x^2}_3)^2} + {({x_3}{x_5} - {x_4}^2)^2} + .... \leqslant 0 \\
\ $
Above condition is possible only when each term of the expression is ZERO
We know that if a, b, c are the three terms which are in GP then if b is the geometric mean of a and c then we can say that ${b^2} = ac$
From this we can say that
  ${x_1}{x_3} = {x^2}_2$,${x_2}{x_4} = {x^2}_3$ and ${x_3}{x_4} = {x^2}_4$
Therefore we can say that ${x_1},{x_2},{x_{3,}}....{x_n}$ are in GP.

NOTE: In this problem Geometric mean property is the important property that has to be applied.
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