<s>[INST] How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(x-y)(y-z)(x-z)=16$$
find this follow minimum of the value
$$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$
My  idea: since
$$\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}=\dfrac{x-z}{(x-y)(y-z)}+\dfrac{1}{x-z}$$
so
$$I=\dfrac{(x-z)^2}{16}+\dfrac{1}{x-z}=\dfrac{(x-z)^2}{16}+\dfrac{1}{2(x-z)}+\dfrac{1}{2(x-z)}\ge\dfrac{3}{4}$$
if and only if $(x-z)=2$,so $(x-y)(y-z)=8$
But we know $$(x-z)^2=[[(x-y)+(y-z)]^2\ge 4(x-y)(y-z)$$
so  this is wrong,
Now  I let $x-z=t$ it is clear $t\ge 4$,so
$$\dfrac{t^2}{16}+\frac{1}{t}=f(t)\Longrightarrow f'(t)\ge 0,t\ge 4$$
so
$$f(t)\ge f(4)=\dfrac{5}{4}$$
My Question: I fell my  methods is ugly,I think this problem have other simple methods.Thank you 
 [/INST] You have reached
$$I = \frac{(x-z)^2}{16}+\frac1{x-z} = \frac{(x-z)^2}{16}+\frac4{x-z}+\frac4{x-z}-\frac7{x-z} \ge 3-\frac7{x-z} \ge \frac54$$
The last inequality is because $$x-z = \frac{16}{(x-y)(y-z)} \ge \frac{64}{(x-z)^2} \implies x-z \ge 4$$

P.S. Equality is when $x-y = y - z = 2$
</s>