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

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

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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Oh, 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.

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

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

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

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

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

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

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

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

Alchemista
 one year ago
Best ResponseYou've already chosen the best response.1I'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.

Alchemista
 one year ago
Best ResponseYou've already chosen the best response.1Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.

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

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

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

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

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

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

Alchemista
 one year ago
Best ResponseYou've already chosen the best response.1Lagrange, 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}]

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

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

Alchemista
 one year ago
Best ResponseYou've already chosen the best response.1http://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
 one year ago
Best ResponseYou've already chosen the best response.2then 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

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0The \(\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\) was the fact I was looking for. I kept thinking the inequality was pointing the other direction :/

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

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

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

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

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

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

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

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

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.0Wait...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
 one year ago
Best ResponseYou've already chosen the best response.2where is the flaw in my proof for this?

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

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

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

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

asnaseer
 one year ago
Best ResponseYou've already chosen the best response.2Ah! 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.

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

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

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

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

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