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ParthKohli Group TitleBest 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 :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
@amistre64 @.Sam. ?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Don't we also need the requirement that \(x,y,z\ge1\)?
 one year ago

ParthKohli Group TitleBest 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.
 one year ago

Bad2zBone Group TitleBest ResponseYou've already chosen the best response.0
Are they integers?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
They are anything satisfying \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\)
 one year ago

wmckinely Group TitleBest 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..._
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
@wmckinely I just know the problem: nothing else ._.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
What if you treat it as a minimization/maximization problem.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
that's a very good way to look at it.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Lagrange_multiplier
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
But the biggest problem is how we proceed.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Never mind, lol. I don't understand that mathematics. :D
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Well, that might be a problem to understanding....
 one year ago

Alchemista Group TitleBest 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.
 one year ago

Alchemista Group TitleBest 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.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
*greater or equal
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
That is not what I am able to understand... how?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
Did you use both constraints when plugging it into wolfram?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
I may have done it a bit wrong though. Let me double check.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
Use the Minimize and Maximize routines.
 one year ago

asnaseer Group TitleBest 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.
 one year ago

Alchemista Group TitleBest 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}]
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
I'll give you the link on wolfram alpha in a moment.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]
 one year ago

Alchemista Group TitleBest 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
 one year ago

asnaseer Group TitleBest 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
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
Anyways, the solution using multivariable calculus is perfectly valid and works.
 one year ago

KingGeorge Group TitleBest 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 :/
 one year ago

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

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

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
I /think/ this works
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Looks good to me.
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
I dont understand that problem,Parth. Obviously.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
thx @KingGeorge :)
 one year ago

help123please. Group TitleBest ResponseYou've already chosen the best response.0
Hi,asnasser. Greetings,George.
 one year ago

Alchemista Group TitleBest 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.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
how is \[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]
 one year ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
assuming a and b are both positive
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
hope that helps
 one year ago

asnaseer Group TitleBest 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.
 one year ago

KingGeorge Group TitleBest 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 \(\sqrt{a}+\sqrt{b}=3+4=7>5\).
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
where is the flaw in my proof for this?
 one year ago

experimentX Group TitleBest 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
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
D'oh! thank for clarifying
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
but does that mean that I have proved the opposite relation to that being asked for?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
no ... \[ \sqrt {x1}+... \ge \sqrt{x1 + ... } \le \sqrt{x+y+z}\] condition is not strong enough
 one year ago

asnaseer Group TitleBest 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.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
Why is no one satisfied with treating this as an optimization problem?
 one year ago

asnaseer Group TitleBest 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. :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Lagrange definitely gives solution but I am still looking for more elementary proof.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
There's just something more appealing about the elementary proofs.
 one year ago

Alchemista Group TitleBest ResponseYou've already chosen the best response.1
I suppose so. At the very least it would be more accessible.
 one year ago
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