How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value

Question

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

You can rephrase it as: Minimize $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for real $a,b,c$ so that $abc=16$ and $a+b+c=0$. — Thomas Andrews, Aug 22 '14 at 16:21

score 5 · Answer 1 · answered Aug 22 '14 at 16:21

5

Let $\displaystyle a=x-y,b=y-z, c=x-z\implies abc=16$ and $\displaystyle a+b-c=0\iff c=a+b$

$$\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}=\dfrac{\dfrac{16}c+c(a+b)}{16}=\dfrac{\dfrac{16}c+c(c)}{16}$$

Now use Second Derivative Test

answered Aug 22 '14 at 16:21

lab bhattacharjee

274,582

It's simple than me! Thank you +1 – math110 Aug 22 '14 at 16:25
1

@shadow That's because $c=2 \implies a+b=2, ab = 8$ which has no solution in real numbers! This answer is NOT correct. – Macavity Aug 22 '14 at 17:51
Can you finish the solution and find the $x, y, z$? – Macavity Aug 22 '14 at 17:52
1

@lab bhattacharjee As pointed out in the solution, I think you need to include the condition that $c\ge4$. – user84413 Aug 22 '14 at 22:05
@user84413, I think that should be dictated by the $2$nd derivative test – lab bhattacharjee Aug 23 '14 at 06:50
Yup, I see that, sorry about that. Unnecessary comment deleted. – shadow10 Aug 23 '14 at 07:34
1

@shadow10 I'm sorry, but I don't see how to get $c=4$ using the 2nd derivative test. Isn't $c=2$ the only critical point? – user84413 Aug 24 '14 at 17:15

Macavity · Answer 2 · 2014-08-22T19:13:51.710

1

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 $x-y = y - z = 2$

edited Aug 22 '14 at 19:13

answered Aug 22 '14 at 18:05

Macavity

46,381

score 0 · Answer 3 · answered Aug 22 '14 at 19:38

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, $$ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {1}{2} + \frac {1}{2} + \frac {1}{4} = \boxed {\dfrac {5}{4}}. $$

How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value

3 Answers3