14. For all real numbers $r, s, t$ satisfying $1 \leqslant r \leqslant s \leqslant t \leqslant 4$, find the minimum value of $(r-1)^{2}+\left(\frac{s}{r}-1\right)^{2}$ $+\left(\frac{t}{s}-1\right)^{2}+\left(\frac{4}{t}-1\right)^{2}$.
14. Let $f(x)=\left(\frac{x}{a}-1\right)^{2}+\left(\frac{b}{x}-1\right)^{2}(0<a \leqslant b, a \leqslant x \leqslant b)$, since $\frac{x}{a}+\frac{b}{x} \geqslant$ $2 \sqrt{\frac{b}{a}} \geqslant 2$, with equality holding if and only if $x=\sqrt{a b}$. Therefore, when $x=\sqrt{a b}$,
$$
f(x)=\left(\frac{x}{a}+\frac{b}{x}-1\right)^{2}+1-\frac{2 b}{a} \geqslant 2\left(\sqrt{\frac{b}{a}}-1\right)^{2} .
$$

It can be seen that, when and only when $r=\sqrt{s}$, $(r-1)^{2}+\left(\frac{s}{r}-1\right)^{2} \geqslant 2(\sqrt{s}-1)^{2}$; when and only when $t=2 \sqrt{s}$, $\left(\frac{t}{s}-1\right)^{2}+\left(\frac{4}{t}-1\right)^{2} \geqslant 2\left(\frac{2}{\sqrt{s}}-1\right)^{2}$; when and only when $\sqrt{s}=\sqrt{2}$, $(\sqrt{s}-1)^{2}+\left(\frac{2}{\sqrt{s}}-1\right)^{2} \geqslant$ $2(\sqrt{2}-1)^{2}$. Thus, when $r=\sqrt{2}, s=2, t=2 \sqrt{2}$, the original expression achieves its minimum value $4(\sqrt{2}-1)^{2}$.