Say I have $x_1,...,x_n$ such that $0 \leq x_i \leq 1$ for all $i \in n$ and $\sum_{i=1}^n x_i = 1$. Let $\epsilon > 0$ be some small number. Let $y_1,...,y_n$ be $\epsilon$-close approximations of $x_i,...,x_n$; specifically, $\lvert y_i - x_i\rvert < \epsilon$ for all $i \in n$. Further, $\sum_{i=1}^ny_i=1$ and $0 \leq y_i \leq 1$ for all $i \in n$.
How can we upper bound in terms of $n$ and $\epsilon$ for the following: $$ \Bigg[1 - \bigg( \sum_{i=1}^n \sqrt{x_iy_i} \bigg)^2\Bigg]^{1/2} $$
EDIT:
It's clear (thanks to @RobertIsrael, see below) that if $\epsilon > 2/n$, then the trivial upper bound of $1$ cannot be beat. What is the bound if $0 < \epsilon < 1/n$?
You can't, because it won't be real. $$\sum_{i=1}^n \sqrt{(x_i+\epsilon) x_i} > \sum_{i=1}^n x_i = 1$$ so you're taking the square root of a negative number.
Perhaps you meant $$ \left[ \left(\sum_{i=1}^n \sqrt{(x_i+\epsilon) x_i}\right)^2 - 1\right]^{1/2} $$
EDIT: OK, for the new question, let $$ Q = \left[ 1 - \left(\sum_{i} \sqrt{x_i y_i}\right)^2\right]^{1/2}$$
You want an upper bound $Q \le U$, presumably better than the trivial $U=1$, which amounts to a lower bound
$$ \sum_{i} \sqrt{x_i y_i} \ge \sqrt{1-U^2} > 0$$
But this won't necessarily be possible. Suppose $n$ is even and $\epsilon > 2/n$, take $x_i = 0$ and $y_i = 2/n$ for $i \le n/2$, $x_i = 2/n$ and $y_i = 0$ for $i > n/2$. Then all $\sqrt{x_i y_i} = 0$ and $Q = 1$.