Question

The question

Answer 1

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

Answer 2

i didnt understand ur statement

Answer 3

$$
\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?

Answer 4

oh yes , u r correct

Answer 5

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}}\)

Answer 6

why did you block me???

Answer 7

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.

Answer 8

well actually idk calculus

Answer 9

HI!!
what class is this?

Answer 10

this is type of question of SAT

Answer 11

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

Answer 12

\[a^2+b^2\ge 2ab\]

Answer 13

\[(x+y+z))(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\] 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 \(z\)

Answer 14

and they are all positive numbers, not negatives allowed

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

\[\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}\]

Answer 20

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?

Answer 21

thnx