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ParthKohliBest ResponseYou've already chosen the best response.1
https://fbcdnsphotosea.akamaihd.net/hphotosakash4/389738_558445504198872_734761968_n.jpg Since I am too lazy to write everything down :)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
@amistre64 @.Sam. ?
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Don't we also need the requirement that \(x,y,z\ge1\)?
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
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.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
They are anything satisfying \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\)
 10 months ago

wmckinelyBest ResponseYou've already chosen the best response.1
Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here..._
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
@wmckinely I just know the problem: nothing else ._.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
What if you treat it as a minimization/maximization problem.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
that's a very good way to look at it.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Lagrange_multiplier
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
But the biggest problem is how we proceed.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Never mind, lol. I don't understand that mathematics. :D
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Well, that might be a problem to understanding....
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
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.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
That is not what I am able to understand... how?
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
Did you use both constraints when plugging it into wolfram?
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
I may have done it a bit wrong though. Let me double check.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
Use the Minimize and Maximize routines.
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
I /think/ you can do this if you use:\[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\]which I believe can be proved.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
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}]
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
I'll give you the link on wolfram alpha in a moment.
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
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
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
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
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
Anyways, the solution using multivariable calculus is perfectly valid and works.
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
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 :/
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
rearranging slightly we get:\[(x+y+z)^2=(x1)^2+2x1+(y1)^2+2y1+(z1)^2+2z1+4xyz\]\[=(x1)^2+(y1)^2+(z1)^2+2x+2y+2z+4xyz3\]
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
then take square roots of both sides to get:\[x+y+z=\sqrt{(x1)^2+(y1)^2+(z1)^2+\text{other terms from above}}\]\[\ge\sqrt{(x1)^2}+\sqrt{(y1)^2}+\sqrt{z1)^2}\]\[\ge(x1)+(y1)+(z1)\]
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
then square root again to get:\[\sqrt{x+y+z}\ge\sqrt{(x1)+(y1)+(z1)}\]\[\ge\sqrt{x1}+\sqrt{y1}+\sqrt{z1}\]
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
I /think/ this works
 10 months ago

help123please.Best ResponseYou've already chosen the best response.0
I dont understand that problem,Parth. Obviously.
 10 months ago

help123please.Best ResponseYou've already chosen the best response.0
Hi,asnasser. Greetings,George.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
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.
 10 months ago

experimentXBest ResponseYou've already chosen the best response.1
how is \[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
\[\begin{align} (\sqrt{a}+\sqrt{b})^2&=a+b+2\sqrt{ab}\\ \therefore a+b&=(\sqrt{a}+\sqrt{b})^22\sqrt{ab}\\ &\ge(\sqrt{a}+\sqrt{b})^2\\ \therefore \sqrt{a+b}&\ge\sqrt{a}+\sqrt{b} \end{align}\]
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
assuming a and b are both positive
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
it may not be a /strict/ proof but it is one that I used to convince myself that this relation might be valid.
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
Wait...I'm no longer convinced. Let \(a=9\), \(b=16\). Then \(\sqrt{a+b}=5\) and \(\sqrt{a}+\sqrt{b}=3+4=7>5\).
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
where is the flaw in my proof for this?
 10 months ago

experimentXBest ResponseYou've already chosen the best response.1
there is a slight error \[ (\sqrt a + \sqrt b)^2  2 \sqrt a \sqrt b \leq (\sqrt a + \sqrt b)^2 \] proves the opposite
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
D'oh! thank for clarifying
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
but does that mean that I have proved the opposite relation to that being asked for?
 10 months ago

experimentXBest ResponseYou've already chosen the best response.1
no ... \[ \sqrt {x1}+... \ge \sqrt{x1 + ... } \le \sqrt{x+y+z}\] condition is not strong enough
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
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.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
Why is no one satisfied with treating this as an optimization problem?
 10 months ago

asnaseerBest ResponseYou've already chosen the best response.2
It's not that I wasn't satisfied  I just wondered if there was a simpler way of arriving at the solution. :)
 10 months ago

experimentXBest ResponseYou've already chosen the best response.1
Lagrange definitely gives solution but I am still looking for more elementary proof.
 10 months ago

KingGeorgeBest ResponseYou've already chosen the best response.0
There's just something more appealing about the elementary proofs.
 10 months ago

AlchemistaBest ResponseYou've already chosen the best response.1
I suppose so. At the very least it would be more accessible.
 10 months ago
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