11. (20 points) Let $x, y, z > 0$, and $xyz = 8$.
(1) If $x + y = 2 \sqrt{\frac{xy}{xy + 8}}$;
$$
(2) Let $\lceil x \rceil$ denote the smallest integer not less than the real number $x$, and $\sum$ denote the cyclic sum, find the value of $\left\lceil \sum \frac{1}{\sqrt{1 + x}} \right\rceil$.
11. (1) Notice,
$$
\frac{x}{1+x}+\frac{y}{1+y} \geqslant 2 \sqrt{\frac{x y}{(1+x)(1+y)}} \text {. }
$$

Thus, it suffices to prove
$$
x y+8>(1+x)(1+y) \Rightarrow x+y\sum \frac{1}{1+x} \\
& =\frac{3+2 \sum x+\sum x y}{(1+x)(1+y)(1+z)} \\
& \geqslant \frac{1+\sum x+\sum x y+x y z}{(1+x)(1+y)(1+z)}=1 .
\end{aligned}
$$

By the symmetry of \(x, y, z\), we may assume \(z \geqslant x \geqslant y\).
Then \(0<y \leqslant 2\).
(i) When \(x+y \geqslant 7\), \(x \geqslant 5\).

Thus, \(z \geqslant x \geqslant 5\). At this time,
$$
u \leqslant \frac{2}{\sqrt{1+5}}+1<2 \text {. }
$$
(ii) When \(x+y<7, \frac{1}{\sqrt{1+z}}=\sqrt{\frac{x y}{x y+8}}\),

By \(\frac{1}{1+y}<\left(\frac{2+y}{2(1+y)}\right)^{2}=\left(1-\frac{y}{2(1+y)}\right)^{2}\)
\(\Rightarrow \frac{1}{\sqrt{1+y}}<1-\frac{y}{2(1+y)}\).
Similarly, \(\frac{1}{\sqrt{1+x}}<1-\frac{x}{2(1+x)}\).
Then \(u<2-\frac{1}{2}\left(\frac{x}{1+x}+\frac{y}{1+y}-2 \sqrt{\frac{x y}{x y+8}}\right)\).
By the conclusion of (1), \(u<2\).
Thus, \(1<u<2\).
Therefore, \(\lceil u\rceil=2\).