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, 09:14 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, 08:12 AMOpalg
Two things need clarifying here. First, it should be stated that all the numbers $\displaystyle x_i$ 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 $\displaystyle x_i = 1/2^i$ then the series becomes

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

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

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 $\displaystyle x_i = 1/2^i$ is the unique sequence that minimises the sum $\displaystyle S = \sum_{i=1}^\infty \frac{x_{i-1}^2}{x_i}$.

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

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

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

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

$\displaystyle x_{n+1} = 2x_n^3/x_{n-1}^2$. -----(*)

Now suppose that (*) holds for all n, and $\displaystyle x_0=1$, $\displaystyle x_1=c$. This recurrence relation has the solution

$\displaystyle x_n = 2^{-n}(2c)^{2^n-1}$. -----(**)

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

If $\displaystyle x_n$ is given by (**) then the series $\displaystyle S$ becomes $\displaystyle \sum_{n=1}^\infty \frac1{2^{n-1}c}$. This is a geometric series with sum $\displaystyle 2/c$. Since $\displaystyle c\leqslant1/2$, the minimum value of the sum is 4.