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mathmath333

  • one year ago

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  1. ybarrap
    • one year ago
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    There are are 2 1/x terms in the second product, that should be 1/z right?

  2. mathmath333
    • one year ago
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    i didnt understand ur statement

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

  4. mathmath333
    • one year ago
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    oh yes , u r correct

  5. mathmath333
    • one year ago
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    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
    • one year ago
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    why did you block me???

  7. ybarrap
    • one year ago
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    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
    • one year ago
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    well actually idk calculus

  9. misty1212
    • one year ago
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    HI!! what class is this?

  10. mathmath333
    • one year ago
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    this is type of question of SAT

  11. misty1212
    • one year ago
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    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
    • one year ago
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    \[a^2+b^2\ge 2ab\]

  13. misty1212
    • one year ago
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    \[(x+y+z))(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\] 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
    • one year ago
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    and they are all positive numbers, not negatives allowed

  15. misty1212
    • one year ago
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    so since you can't tell the difference between the numbers, it will have a minimum when they are all equal

  16. misty1212
    • one year ago
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    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
    • one year ago
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    like i said, it is just thinking, your math teacher might have a more complicated explanation

  18. ganeshie8
    • one year ago
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    I'm not so sure about that symmetry argument but we can use AM-GM inequality

  19. ganeshie8
    • one year ago
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    \[\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right) &= \frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{z^2+x^2}{zx}+3 \\~\\ &\ge \frac{2xy}{xy}+\frac{2yz}{yz}+\frac{2zx}{zx}+3\\~\\ &\ge 2+2+2+3\\~\\ \end{align}\]

  20. misty1212
    • one year ago
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    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
    • one year ago
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    thnx

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