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Given these condition, I am seeking a proof:

Define a sequence of real numbers by $x_1 = 3$ and then, for $n \geq 2$,

$x_n = \sqrt{2 \; x_{n-1} + 1 }$

Prove that for all positive integers $n$, $x_n \geq x_{n+1}$.

I began a proof by induction, but ran out of steam. I tested the base case for $n=2$ , but I could not seem to get anywhere after that. I feel like there is not enough information (ie recursion, sequence, etc) to prove by induction. Is an induction proof an efficient way to proceed? Are there easier methods of proof?

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    Aryabhata is a veteran here, it is indeed a duplicate.I have been on this group only past 23 days and have to learn every day. (I am his fan!)2012-03-16

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For the induction, assume that $x_n\leq x_{n-1}$. Then

$ \begin{align*} x_n &\leq x_{n-1}\\ 2x_n &\leq 2x_{n-1}\\ 2x_n+1 &\leq 2x_{n-1}+1\\ \sqrt{2x_n+1} &\leq \sqrt{2x_{n-1}+1}\\ \end{align*} $

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    The last line is exactly $x_{n+1}\leq x_n$.2012-03-16
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$ x_n \geq x_{n+1} $ if and only if $x_n \geq \sqrt{2x_n+1}$ and this is true if and only if ${x_n}^2 \geq 2x_n+1$ which is same as ${x_n}^2 -2x_n-1 \geq 0$, and this expression can be written as $(x_n+\sqrt{2}-1)(x_n-\sqrt{2}-1) \geq 0$

The signs in first expression was flipped, just fixed it (Patrick was right)

So try to show that expression is always positive for every $n$

Now how do we show $x_n \geq \sqrt{2}+1$ and $x_n \geq 1-\sqrt{2}$ for every $n$ by induction?

I think Joe Johnson has better proof, follow that.

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    But my expression is different from kirthi's thats for sure.2012-03-16
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Calculate $x_2$, and notice that $x_{n+1}-x_n$ has the same sign as $x_{n}-x_{n-1}$, and inductively, the same sign as $x_2-x_1$, which you just calculated.