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ParthKohli
 3 years ago
Can anyone help me with this one?
ParthKohli
 3 years ago
Can anyone help me with this one?

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ParthKohli
 3 years 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 :)

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here..._

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What if you treat it as a minimization/maximization problem.

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years ago
Best ResponseYou've already chosen the best response.0Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.

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

KingGeorge
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you use both constraints when plugging it into wolfram?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use the Minimize and Maximize routines.

asnaseer
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lagrange, 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'll give you the link on wolfram alpha in a moment.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://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
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyways, the solution using multivariable calculus is perfectly valid and works.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I dont understand that problem,Parth. Obviously.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hi,asnasser. Greetings,George.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In 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
 3 years ago
Best ResponseYou've already chosen the best response.1how is \[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]

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

asnaseer
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2where is the flaw in my proof for this?

experimentX
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.2D'oh! thank for clarifying

asnaseer
 3 years 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
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why is no one satisfied with treating this as an optimization problem?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I suppose so. At the very least it would be more accessible.
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