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I have encountered the following problem and I am curious how to solve it.

$\textrm{Given }a_{n+1} = a_{n}(1 - \sqrt{a_{n}}) \textrm{, where } a_{i} \in (0,1) \textrm{, } i = \overline{1,n}$

I have proved that $(a_{n})_{n\in\mathbb{N}}$ is decreasing and now I have to prove that the upper bound of

$b_{n} = {a_{1}^2} + {a_{2}^2} + \cdots + {a_{n}^2}$ is $a_{1}$.

I have no idea how to do this. I have tried in all sorts of ways but only get to something like

$b_{n} < {n}\cdot{a_{1}^2}$ or $b_{n} < {n}\cdot{(1-\sqrt{a_{1}})^2}$

which is not even close to what the upper bound must be. I feel like it's a common trick that you have to use to solve this, but I cannot find it.

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Note that since $a_n>0$ for any $n$ we have $$ a_{n+1} = a_n(1-\sqrt{a_n})\frac{1+\sqrt{a_n}}{1+\sqrt{a_n}} = \frac{a_n(1-a_n)}{1+\sqrt{a_n}} < a_n(1-a_n) = a_n - a_{n}^{2}. $$ Therefore we get $$ a_1 > a_2 + a_{1}^{2} > \ldots> a_n + a_{n}^2 + \ldots + a_{1}^{2} > a_{n}^2 + \ldots + a_{1}^{2}. $$

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  • $\begingroup$ Nicely done. Thanks. (+1) $\endgroup$ – Kay K. Dec 12 '15 at 2:43
  • $\begingroup$ Thanks very much. This is so elegant. $\endgroup$ – micha987123 Dec 12 '15 at 10:18
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We induct on $n$. This is clearly true when $n=1$, as $a_1^2\leq a_1$.

Suppose this is true for a sequence of length $n$. Let's show this for a sequence of length $n+1$.

We notice that $a_1^2+a_2^2+a_3^2+\dots+a_n^2+a_{n+1}^2=a_1^2+\left(a_2^2+a_3^2+\dots+a_n^2+a_{n+1}^2\right)$, where the bracketed part is a sequence of length $n$.

$a_2^2+a_3^2\dots+a_{n+1}^2$ is a sequence of length $n$, if we relabel $a_i$ as $a_{i-1}$, it would look like $a_1^2+a_2^2\dots+a_n^2$. This can be done since the sequence is defined recursively, with $a_{i+1}$ depending solely on $a_i$ the same way $a_{i+2}$ depends on $a_{i+1}$, it is independent of the position of the sequence $a_i$ is.

By induction hypothesis, we have $\left(a_2^2+a_3^2+\dots+a_n^2+a_{n+1}^2\right)\leq a_2$.

Hence it suffices to show that $a_1^2+a_2\leq a_1$.

Substituting the value of $a_2$, we have $a_1^2+a_1\left(1-\sqrt{a_1}\right)\leq a_1$.

This reduces to $a_1^2\leq a_1\sqrt{a_1}$, which is true as $a_1\in(0,1)$.

By mathematical induction, this is true for all $n$.

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  • $\begingroup$ Isn't the thing you want to prove "$a_1^2+a_2^2+a_3^2+\dots+a_n^2 \le a_1$"? Then is your "induction hypothesis" for $n$ correct? $\endgroup$ – Kay K. Dec 12 '15 at 2:31
  • $\begingroup$ $a_2^2+\dots+a_{n+1}^2$ is a sequence of length $n$, if we relabel $a_i$ as $a_{i-1}$, it would look like $a_1^2+\dots+a_n^2$. $\endgroup$ – Element118 Dec 12 '15 at 2:34
  • $\begingroup$ Ha. Very interesting! (+1) $\endgroup$ – Kay K. Dec 12 '15 at 2:42

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