Trouble finding the upper bound for a certain sum. 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.
 A: 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
$$
a_1 > a_2 + a_{1}^{2} > \ldots> a_n + a_{n}^2 + \ldots + a_{1}^{2} > a_{n}^2 + \ldots + a_{1}^{2}.
$$
A: 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 $n$.
