How can $x_{n+1}=1-\sqrt{1-x_n}$ be a decreasing sequence on [0, 1)? For $x_{n+1}=1-\sqrt{1-x_n}$ on $[0, 1)$, we have that $(1-x_{n+1})^2=1-x_n$, where LHS and RHS are both less than $1$, which implies that the sequence is increasing. But the textbook asks to prove that it's decreasing. What's my error?
 A: Assuming the sequence in non-negative (as you proved or were given. It appears in the question) 
$$x_{n+1}\ge x_n\iff 1-\sqrt{1-x_n}\ge x_n\iff 1-2x_n+x^2_n\ge1-x_n\iff$$
$$\iff x_n^2-x_n\ge0\iff x_n(x_n-1)\le0$$
and the last inequality is trivial now...and I also get the sequence is decreasing.
A: There is a simpler approach :
Consider the auxiliary sequence $v_n=1-u_n$ with $v_n\in [0,1]$, that verifies $v_{n+1}=\sqrt{v_n}$ : it is increasing to $1$; thus $u_n$ is  decreasing to $0$.
(many thanks to @Lutzl for correcting me) 
A: One may write, for $n\geq1$,

$$
\begin{align}
x_n-x_{n+1}&=\left(1-\sqrt{1-x_{n-1}}\right)-\left(1-\sqrt{1-x_n}\right)\\\\
&=\sqrt{1-x_n}-\sqrt{1-x_{n-1}}\\\\
&=\frac{x_{n-1}-x_n}{\sqrt{1-x_n}+\sqrt{1-x_{n-1}}}\\\\
&=\cdots\\\\
&=\frac{x_0-x_1}{\prod_{k=1}^n\left(\sqrt{1-x_k}+\sqrt{1-x_{k-1}}\right)}\:\color{blue}{\geq0}
\end{align}
$$ 

since $x_0-x_1=\sqrt{1-x_0}\:(1-\sqrt{1-x_0})$.
A: For example, let $x_n=\frac{9}{25}$. Then $x_{n+1}=1-\sqrt{1-9/25}=1/5\lt x_n$. In general,
$$x_{n+1}=(1-\sqrt{1-x_n})\cdot \frac{1+\sqrt{1-x_n}}{1+\sqrt{1-x_n}}=  \frac{x_n}{1+\sqrt{1-x_n}}\lt x_n.$$
The sequence is indeed decreasing.
