Can anyone help me with this one?

- ParthKohli

Can anyone help me with this one?

- jamiebookeater

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- ParthKohli

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

- ParthKohli

@amistre64 @.Sam. ?

- KingGeorge

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

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## More answers

- ParthKohli

Yes.

- ParthKohli

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.

- anonymous

Are they integers?

- ParthKohli

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

- anonymous

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

- ParthKohli

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

- anonymous

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

- ParthKohli

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

- anonymous

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

- ParthKohli

But the biggest problem is how we proceed.

- ParthKohli

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

- KingGeorge

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

- anonymous

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.

- anonymous

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

- anonymous

*greater or equal

- ParthKohli

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

- KingGeorge

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

- anonymous

Ah ok

- anonymous

Did you use both constraints when plugging it into wolfram?

- KingGeorge

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

- anonymous

Use the Minimize and Maximize routines.

- asnaseer

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

- anonymous

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}]

- anonymous

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

- asnaseer

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

- anonymous

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

- asnaseer

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

- anonymous

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

- KingGeorge

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 :/

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

I /think/ this works

- KingGeorge

Looks good to me.

- anonymous

I dont understand that problem,Parth.
Obviously.

- asnaseer

thx @KingGeorge :)

- anonymous

Hi,asnasser. Greetings,George.

- anonymous

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.

- experimentX

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

- asnaseer

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

- asnaseer

assuming a and b are both positive

- asnaseer

hope that helps

- asnaseer

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

- KingGeorge

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

- asnaseer

:(

- asnaseer

where is the flaw in my proof for this?

- experimentX

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

- asnaseer

D'oh! thank for clarifying

- asnaseer

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

- experimentX

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

- asnaseer

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.

- anonymous

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

- asnaseer

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

- experimentX

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

- KingGeorge

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

- anonymous

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

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