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Math Help - prove problem 2

  1. #1
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    prove problem 2

    For real numbers x[i] such that x[0]=1, x[i+1]<=x[i] for i=1, 2,...,n.

    prove that

    x[0]^2/x[1] + x[1]^2/x[2] + x[2]^2/x[3] + ... + x[n-1]^2/x[n] >= 4

    can anyone solve this..? thank you
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  2. #2
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    Quote Originally Posted by nzm88 View Post
    For real numbers x_i such that x_0=1,\ x_{i+1}\leqslant x_i for i=1, 2,...,n, prove that

    \frac{x_0^2}{x_1} + \frac{x_1^2}{x_2} + \frac{x_2^2}{x_3} + \ldots + \frac{x_{n-1}^2}{x_n} \geqslant 4.
    Two things need clarifying here. First, it should be stated that all the numbers x_i are positive. Second, the series must be an infinite sum. If it allowed to stop after finitely many terms then the result is false. For example, if x_i = 1/2^i then the series becomes

    2 + 1 + \tfrac12 + \tfrac14 + \tfrac18 + \ldots.

    If that series stops after finitely many terms then the sum will be less than 4. But the sum of the infinite series is equal to 4.

    So the problem should be stated as follows:

    For positive numbers x_i such that x_0=1,\ x_{i+1}\leqslant x_i for i\geqslant1, prove that \sum_{i=1}^\infty \frac{x_{i-1}^2}{x_i} \geqslant 4.

    With those modifications, this becomes an interesting problem. Here is an outline solution. It's a bit laborious, so maybe someone can come up with a better method. The basic idea is that the above sequence x_i = 1/2^i is the unique sequence that minimises the sum S = \sum_{i=1}^\infty \frac{x_{i-1}^2}{x_i}.

    So we start with a decreasing sequence (x_n) of positive numbers, and we see whether we can modify it so as to reduce the sum S. Notice first that if two terms in the sequence are equal, say x_n=x_{n-1}, then we can reduce S by omitting x_n from the sequence, and going straight from x_{n-1} to x_{n+1}. (You can easily check that this will reduce S by an amount x_{n-1}.)

    Suppose that we keep all the terms fixed except for x_n, and we look for the value of x_n that minimises S. The only terms in S that involve x_n are \frac{x_{n-1}^2}{x_n} and \frac{x_n^2}{x_{n+1}}. Write a=x_{n-1}, x = x_n and b = x_{n+1}. Let f(x) = \frac{a^2}x + \frac{x^2}b. Then we look for the minimum value of f(x) in the interval [b,a].

    You can check by calculus that if b\geqslant a/\sqrt2 then the minimum occurs at the endpoint x=b. But that corresponds to the situation x_n = x_{n-1}, and we already saw that in that case we can reduce S by omitting x_n from the sequence.

    So we may assume that b< a/\sqrt2, and that the minimum of f(x) occurs at the turning point x = (a^2b/2)^{1/3} inside the interval [a,b]. Thus we can minimise S by taking x_n = (x_{n-1}^2x_{n+1}/2)^{1/3}, which is equivalent to
    x_{n+1} = 2x_n^3/x_{n-1}^2. -----(*)

    Now suppose that (*) holds for all n, and x_0=1, x_1=c. This recurrence relation has the solution
    x_n = 2^{-n}(2c)^{2^n-1}. -----(**)

    However, we must check again that x_{n+1}<x_n. That gives the inequality 2^{-n-1}(2c)^{2^{n+1}-1} < 2^{-n}(2c)^{2^n-1}, which reduces to (2c)^{2^n}<2. For that to hold for all n implies that 2c\leqslant1, or c\leqslant1/2.

    If x_n is given by (**) then the series S becomes \sum_{n=1}^\infty \frac1{2^{n-1}c}. This is a geometric series with sum 2/c. Since c\leqslant1/2, the minimum value of the sum is 4.
    Last edited by Opalg; August 5th 2009 at 10:54 AM. Reason: corrected typo
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