<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] First, let $ a=x-y $, $b=y-z$, and $c=x-z$. Then, note that $ a + b - c = 0 $; i.e. $c=a+b$ and $a,b,c>0$. Finally, $abc=16$; i.e. $ab(a+b)=16$. Then, $$ I = \frac {1}{a} + \frac {1}{b} + \frac {1}{a+b}. $$You can either finish like lab bhattacharjee did, or you can use the method of Lagrange Multipliers. I will post the Lagrange solution. Note that $$ \nabla I = \left< - \frac {1}{a^2} - \frac {1}{(a+b)^2}, - \frac {1}{b^2} - \frac {1}{(a+b)^2} \right> $$and $$ \nabla \left( \text{constraint} \right) = \nabla \left( ab(a+b) \right) = \left< b(2a+b), a(a+2b) \right>. $$Now, these two gradients must be proportional, so we get $$ \frac {b(2a+b)}{\frac{1}{a^2} + \frac {1}{(a+b)^2}} = \frac {a(a+2b)}{\frac{1}{b^2} + \frac {1}{(a+b)^2}}. $$Doing some algebra yields $ a = b $ when $a,b>0$. Hence, $c=a+b=2a$, so $$ abc = a \cdot a \cdot \left( 2a \right) = 16 \iff a = 2, $$ giving $ b = 2 $ and $ c = 4 $, and the minimum value of $I$ is, therefore, $$ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {1}{2} + \frac {1}{2} + \frac {1}{4} = \boxed {\dfrac {5}{4}}. $$
</s>