## mathmath333 one year ago The question

1. ybarrap

There are are 2 1/x terms in the second product, that should be 1/z right?

2. mathmath333

i didnt understand ur statement

3. ybarrap

\large \color{black}{\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1{\color{red}{x}}\right)\hspace{.33em}\\~\\ \end{align}} should be \large \color{black}{\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right)\hspace{.33em}\\~\\ \end{align}} right?

4. mathmath333

oh yes , u r correct

5. mathmath333

find the minimum value of \large \color{black}{\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right)\hspace{.33em}\\~\\ \end{align}} if \large \color{black}{\begin{align} \{x,y,z\}\in \mathbb{R^{>0}}\hspace{.33em}\\~\\ \end{align}}

6. anonymous

why did you block me???

7. ybarrap

Multiplying everything out we get $$x/y+y/x+x/z+z/x+z/y+y/z+3$$ Using method of lagrange multipliers with objective function $$f(x,y,z)=x/y+y/x+x/z+z/x+z/y+y/z+3$$ and constraint $$xyz=0$$ We get $$x/y=1\\ x/z=1\\ y/z=1\\$$ So minimum is $$x/y+y/x+x/z+z/x+z/y+y/z+3=6\times 1+3=9$$ Are you familiar with this process? https://en.wikipedia.org/wiki/Lagrange_multiplier Note, although x>0,y>0,z>0, I used xyz=0 for the constraint since this is the lower bound of their domain. Here is the setup $$\Lambda(x,y,z)=f(x,y,z)+\lambda(xyz-0)\\$$ The first differential is $$\Lambda_x=1/y-y/x^2+1/z-z/x^2+\lambda yz=0$$ The other differentials are similar Using the constraint and this equation I get x/y=1 The other results follow in a similar manner.

8. mathmath333

well actually idk calculus

9. misty1212

HI!! what class is this?

10. mathmath333

this is type of question of SAT

11. misty1212

just asking because i think we can precede just by thinking, although i am not sure your math teacher will like it q

12. ganeshie8

$a^2+b^2\ge 2ab$

13. misty1212

$(x+y+z))(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ is completely symmetric in $$x,y,z$$ by which i mean if you permute them you get the same thing another way of saying you can't tell the difference between $$x,y$$ and $$z$$

14. misty1212

and they are all positive numbers, not negatives allowed

15. misty1212

so since you can't tell the difference between the numbers, it will have a minimum when they are all equal

16. misty1212

if $$x>1$$ then $$\frac{1}{x}<1$$ to balance it out, make them all 1 and you get $(1+1+1)(1+1+1)=9$

17. misty1212

like i said, it is just thinking, your math teacher might have a more complicated explanation

18. ganeshie8

I'm not so sure about that symmetry argument but we can use AM-GM inequality

19. ganeshie8

\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right) &= \frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{z^2+x^2}{zx}+3 \\~\\ &\ge \frac{2xy}{xy}+\frac{2yz}{yz}+\frac{2zx}{zx}+3\\~\\ &\ge 2+2+2+3\\~\\ \end{align}

20. misty1212

i am not either, but not only is it symmetric in x, y, z, it is also symmetric in 1/x,1/y,1/z so how could it be anything other than x = y = z = 1?

21. mathmath333

thnx