11. Real numbers $r, s, t$ satisfy $1 \leqslant r \leqslant s \leqslant t \leqslant 4$, then $(r-1)^{2}+\left(\frac{s}{r}-1\right)^{2}+\left(\frac{t}{s}-1\right)^{2}+\left(\frac{4}{t}-1\right)^{2}$ has the minimum value of . $\qquad$
11. $12-8 \sqrt{2}$ Since $(x-1)^{2}-((2 \sqrt{2}-2) x-1)=(x-\sqrt{2})^{2} \geqslant 0$, we have $(x-1)^{2} \geqslant(2 \sqrt{2}-2) x-$
1. Substituting $x=r, \frac{s}{r}, \frac{t}{s}, \frac{4}{t}$ into the above inequality and adding them up, we get
$$
\begin{aligned}
(r-1)^{2}+\left(\frac{s}{r}-1\right)^{2}+\left(\frac{t}{s}-1\right)^{2}+\left(\frac{4}{t}-1\right)^{2} & \geqslant(2 \sqrt{2}-2)\left(r+\frac{s}{r}+\frac{t}{s}+\frac{4}{t}\right)-4 \\
& \geqslant(2 \sqrt{2}-2) \cdot 4 \sqrt{2}-4=12-8 \sqrt{2},
\end{aligned}
$$

Equality holds when $r=\sqrt{2}, s=2, t=2 \sqrt{2}$.