Question

Can anyone help me with this one?

Answer 1

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash4/389738_558445504198872_734761968_n.jpg Since I am too lazy to write everything down :-)

Answer 2

@amistre64 @.Sam. ?

Answer 3

Don't we also need the requirement that \(x,y,z\ge1\)?

Answer 4

Yes.

Answer 5

Oh, and it's not a homework or a test... just found it and wondered how to solve it. If you know the solution, then please solve it and show complete steps.

Answer 6

Are they integers?

Answer 7

They are anything satisfying \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\)

Answer 8

Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here...-_-

Answer 9

@wmckinely I just know the problem: nothing else ._.

Answer 10

What if you treat it as a minimization/maximization problem.

Answer 11

that's a very good way to look at it.

Answer 12

http://en.wikipedia.org/wiki/Lagrange_multiplier

Answer 13

But the biggest problem is how we proceed.

Answer 14

Never mind, lol. I don't understand that mathematics. :-D

Answer 15

Well, that might be a problem to understanding....

Answer 16

I'd have to actually do the math. But it looks like a simple application of the processed I linked you to. You have an equality constraint and function both over three variables.

Answer 17

Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.

Answer 18

*greater or equal

Answer 19

That is not what I am able to understand... how?

Answer 20

Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.

Answer 21

Ah ok

Answer 22

Did you use both constraints when plugging it into wolfram?

Answer 23

I may have done it a bit wrong though. Let me double check.

Answer 24

Use the Minimize and Maximize routines.

Answer 25

I /think/ you can do this if you use:\[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\]which I believe can be proved.

Answer 26

Lagrange, worked:
Minimize[{Sqrt[x^2 + y^2 + z^2], 1/x + 1/y + 1/z == 2, x > 0, y > 0,
z > 0}, {x, y, z}] = Maximize[{Sqrt[x - 1] + Sqrt[y - 1] + Sqrt[z - 1],
1/x + 1/y + 1/z == 2, x > 0, y > 0, z > 0 }, {x, y, z}]

Answer 27

I'll give you the link on wolfram alpha in a moment.

Answer 28

if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]

Answer 29

http://www.wolframalpha.com/input/?i=Minimize%5B%7BSqrt%5Bx%5E2+%2B+y%5E2+%2B+z%5E2%5D%2C+1%2Fx+%2B+1%2Fy+%2B+1%2Fz+%3D%3D+2%2C+x+%3E+0%2C+y+%3E+0%2C++++z+%3E+0%7D%2C+%7Bx%2C+y%2C+z%7D%5D
http://www.wolframalpha.com/input/?i=Maximize%5B%7BSqrt%5Bx+-+1%5D+%2B+Sqrt%5By+-+1%5D+%2B+Sqrt%5Bz+-+1%5D%2C++++1%2Fx+%2B+1%2Fy+%2B+1%2Fz+%3D%3D+2%2C+x+%3E+0%2C+y+%3E+0%2C+z+%3E+0+%7D%2C+%7Bx%2C+y%2C+z%7D%5D

Answer 30

then note that:\[(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=x^2+y^2+z^2+4xyz\] from above

Answer 31

Anyways, the solution using multivariable calculus is perfectly valid and works.

Answer 32

The \(\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\) was the fact I was looking for. I kept thinking the inequality was pointing the other direction :/

Answer 33

rearranging slightly we get:\[(x+y+z)^2=(x-1)^2+2x-1+(y-1)^2+2y-1+(z-1)^2+2z-1+4xyz\]\[=(x-1)^2+(y-1)^2+(z-1)^2+2x+2y+2z+4xyz-3\]

Answer 34

then take square roots of both sides to get:\[x+y+z=\sqrt{(x-1)^2+(y-1)^2+(z-1)^2+\text{other terms from above}}\]\[\ge\sqrt{(x-1)^2}+\sqrt{(y-1)^2}+\sqrt{z-1)^2}\]\[\ge(x-1)+(y-1)+(z-1)\]

Answer 35

then square root again to get:\[\sqrt{x+y+z}\ge\sqrt{(x-1)+(y-1)+(z-1)}\]\[\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\]

Answer 36

I /think/ this works

Answer 37

Looks good to me.

Answer 38

I dont understand that problem,Parth.
Obviously.

Answer 39

thx @KingGeorge :)

Answer 40

Hi,asnasser. Greetings,George.

Answer 41

In any case you will notice that Lagrange finds the local min/max at (1.5, 1.5, 1.5) for those functions subject to the given constraints.

Answer 42

how is
\[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]

Answer 43

\[\begin{align}
(\sqrt{a}+\sqrt{b})^2&=a+b+2\sqrt{ab}\\
\therefore a+b&=(\sqrt{a}+\sqrt{b})^2-2\sqrt{ab}\\
&\ge(\sqrt{a}+\sqrt{b})^2\\
\therefore \sqrt{a+b}&\ge\sqrt{a}+\sqrt{b}
\end{align}\]

Answer 44

assuming a and b are both positive

Answer 45

hope that helps

Answer 46

it may not be a /strict/ proof but it is one that I used to convince myself that this relation might be valid.

Answer 47

Wait...I'm no longer convinced. Let \(a=9\), \(b=16\). Then \(\sqrt{a+b}=5\) and \(\sqrt{a}+\sqrt{b}=3+4=7>5\).

Answer 48

:(

Answer 49

where is the flaw in my proof for this?

Answer 50

there is a slight error
\[ (\sqrt a + \sqrt b)^2 - 2 \sqrt a \sqrt b \leq (\sqrt a + \sqrt b)^2 \]
proves the opposite

Answer 51

D'oh! thank for clarifying

Answer 52

but does that mean that I have proved the opposite relation to that being asked for?

Answer 53

no ...
\[ \sqrt {x-1}+... \ge \sqrt{x-1 + ... } \le \sqrt{x+y+z}\]
condition is not strong enough

Answer 54

Ah! I see now - thanks again @experimentX (and @KingGeorge)
Guess I'll have to think again on this problem - maybe calculas is the way to go as you guys showed above.

Answer 55

Why is no one satisfied with treating this as an optimization problem?

Answer 56

It's not that I wasn't satisfied - I just wondered if there was a simpler way of arriving at the solution. :)

Answer 57

Lagrange definitely gives solution but I am still looking for more elementary proof.

Answer 58

There's just something more appealing about the elementary proofs.

Answer 59

I suppose so. At the very least it would be more accessible.