14. Make the linear substitution $\left\{\begin{array}{l}x=b+2 c+3 d \\ y=c+2 d+3 a \\ z=d+2 a+3 b \\ w=a+2 b+3 c\end{array}\right.$, treating $a, b, c, d$ as variables, and solve to get
$$\left\{\begin{array}{l} a=-\frac{5}{24} x+\frac{7}{24} y+\frac{1}{24} z+\frac{1}{24} w \\ b=\frac{1}{24} x-\frac{5}{24} y+\frac{7}{24} z+\frac{1}{24} w \\ c=\frac{1}{24} x+\frac{1}{24} y-\frac{5}{24} z+\frac{7}{24} w \\ d=\frac{7}{24} x+\frac{1}{24} y+\frac{1}{24} z-\frac{5}{24} w \end{array}\right.$$

Let the left side of the original inequality be $M$. Substituting the above four equations, we get
$$\begin{aligned} M= & \frac{7}{24}\left(\frac{y}{x}+\frac{z}{y}+\frac{w}{z}+\frac{x}{w}\right)+\frac{1}{24}\left(\frac{z}{x}+\frac{w}{y}+\frac{x}{z}+\frac{y}{w}\right)+ \\ & \frac{1}{24}\left(\frac{w}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{w}\right)-\frac{5}{6} \geqslant \\ & \frac{7}{24}\left(4 \sqrt[4]{\frac{y}{x} \cdot \frac{z}{y} \cdot \frac{w}{z} \cdot \frac{x}{w}}\right)+\frac{1}{24}\left(4 \sqrt[4]{\frac{z}{x} \cdot \frac{w}{y} \cdot \frac{x}{z} \cdot \frac{y}{w}}\right)+ \\ & \frac{1}{24}\left(4 \sqrt[4]{\frac{w}{x} \cdot \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{w}}\right)-\frac{5}{6}=\frac{2}{3} \end{aligned}$$