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(Hi)

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- Aug 4th 2009, 10:14 PMnzm88prove 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(Hi) - Aug 5th 2009, 09:12 AMOpalg
Two things need clarifying here. First, it should be stated that all the numbers 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 then the series becomes

.

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 such that for , prove that .

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 is the unique sequence that minimises the sum .

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

Suppose that we keep all the terms fixed except for , and we look for the value of that minimises . The only terms in that involve are and . Write , and . Let . Then we look for the minimum value of f(x) in the interval [b,a].

You can check by calculus that if then the minimum occurs at the endpoint . But that corresponds to the situation , and we already saw that in that case we can reduce by omitting from the sequence.

So we may assume that , and that the minimum of f(x) occurs at the turning point inside the interval [a,b]. Thus we can minimise by taking , which is equivalent to

. -----(*)

Now suppose that (*) holds for all n, and , . This recurrence relation has the solution

. -----(**)

However, we must check again that . That gives the inequality , which reduces to . For that to hold for all n implies that , or .

If is given by (**) then the series becomes . This is a geometric series with sum . Since , the minimum value of the sum is 4.