Which inequality?

Algebra Level 4

If x 1 , x 2 , x 3 , … , x n x_1,~x_2,~x_3,\dots,x_n are n n non-zero real numbers such ( 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 ) ≤ ( x 1 x 2 + x 2 x 3 + x 3 x 4 + ⋯ + x n − 1 x n ) 2 (x_1^2+x_2^2+x_3^2+\dots+x_{n-1}^2)(x_2^2+x_3^2+x_4^2+\dots+x_n^2)\leq(x_1x_2+x_2x_3+x_3x_4+\dots+x_{n-1}x_n)^2 then x 1 , x 2 , x 3 , … , x n x_1,~x_2,~x_3,\dots,x_n are in

Note :
- AP denotes an arithmetic progression.
- AGP denotes an arithmetic geometric progression.
- GP denotes an geometric progression.
- HP denotes an harmonic progression.

None of the others GP HP AP AGP

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Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Ravi Dwivedi
Jul 13, 2015

Cauchy Schwarz inequality implies ( 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 ) ≥ ( x 1 x 2 + x 2 x 3 + x 3 x 4 + ⋯ + x n − 1 x n ) 2 (x_1^2+x_2^2+x_3^2+\dots+x_{n-1}^2)(x_2^2+x_3^2+x_4^2+\dots+x_n^2)\geq(x_1x_2+x_2x_3+x_3x_4+\dots+x_{n-1}x_n)^2 But the given is

( 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 ) ≤ ( x 1 x 2 + x 2 x 3 + x 3 x 4 + ⋯ + x n − 1 x n ) 2 (x_1^2+x_2^2+x_3^2+\dots+x_{n-1}^2)(x_2^2+x_3^2+x_4^2+\dots+x_n^2)\leq(x_1x_2+x_2x_3+x_3x_4+\dots+x_{n-1}x_n)^2

The only possibility is ( 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 ) = ( x 1 x 2 + x 2 x 3 + x 3 x 4 + ⋯ + x n − 1 x n ) 2 (x_1^2+x_2^2+x_3^2+\dots+x_{n-1}^2)(x_2^2+x_3^2+x_4^2+\dots+x_n^2)=(x_1x_2+x_2x_3+x_3x_4+\dots+x_{n-1}x_n)^2

Equality holds when x 1 x 2 = x 2 x 3 = x 3 x 4 = . . . = x n − 1 x n = r \frac{x_1}{x_2}=\frac{x_2}{x_3}=\frac{x_3}{x_4}=...=\frac{x_{n-1}}{x_n}=r

x 1 , x 2 , . . . x n x_{1},x_{2},...x_{n} are in G.P.

This r r is the common ratio of this G.P.

7 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Good approach. Observing that we have the equality case of an inequality, gives us a ton more information.

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