mathmath333
  • mathmath333
The question
Linear Algebra
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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ybarrap
  • ybarrap
There are are 2 1/x terms in the second product, that should be 1/z right?
mathmath333
  • mathmath333
i didnt understand ur statement
ybarrap
  • 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?

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mathmath333
  • mathmath333
oh yes , u r correct
mathmath333
  • mathmath333
find the minimum value of \(\large \color{black}{\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right)\hspace{.33em}\\~\\ \end{align}}\) if \(\large \color{black}{\begin{align} \{x,y,z\}\in \mathbb{R^{>0}}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
why did you block me???
ybarrap
  • 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.
mathmath333
  • mathmath333
well actually idk calculus
misty1212
  • misty1212
HI!! what class is this?
mathmath333
  • mathmath333
this is type of question of SAT
misty1212
  • misty1212
just asking because i think we can precede just by thinking, although i am not sure your math teacher will like it q
ganeshie8
  • ganeshie8
\[a^2+b^2\ge 2ab\]
misty1212
  • 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 \(z\)
misty1212
  • misty1212
and they are all positive numbers, not negatives allowed
misty1212
  • misty1212
so since you can't tell the difference between the numbers, it will have a minimum when they are all equal
misty1212
  • misty1212
if \(x>1\) then \(\frac{1}{x}<1\) to balance it out, make them all 1 and you get \[(1+1+1)(1+1+1)=9\]
misty1212
  • misty1212
like i said, it is just thinking, your math teacher might have a more complicated explanation
ganeshie8
  • ganeshie8
I'm not so sure about that symmetry argument but we can use AM-GM inequality
ganeshie8
  • ganeshie8
\[\begin{align} (x+y+z)\left(\dfrac1x+\dfrac1y+\dfrac1z\right) &= \frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{z^2+x^2}{zx}+3 \\~\\ &\ge \frac{2xy}{xy}+\frac{2yz}{yz}+\frac{2zx}{zx}+3\\~\\ &\ge 2+2+2+3\\~\\ \end{align}\]
misty1212
  • 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?
mathmath333
  • mathmath333
thnx

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