ParthKohli
  • ParthKohli
Can anyone help me with this one?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ParthKohli
  • 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
  • ParthKohli
@amistre64 @.Sam. ?
KingGeorge
  • KingGeorge
Don't we also need the requirement that \(x,y,z\ge1\)?

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More answers

ParthKohli
  • ParthKohli
Yes.
ParthKohli
  • 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
  • anonymous
Are they integers?
ParthKohli
  • ParthKohli
They are anything satisfying \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\)
anonymous
  • anonymous
Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here...-_-
ParthKohli
  • ParthKohli
@wmckinely I just know the problem: nothing else ._.
anonymous
  • anonymous
What if you treat it as a minimization/maximization problem.
ParthKohli
  • ParthKohli
that's a very good way to look at it.
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Lagrange_multiplier
ParthKohli
  • ParthKohli
But the biggest problem is how we proceed.
ParthKohli
  • ParthKohli
Never mind, lol. I don't understand that mathematics. :-D
KingGeorge
  • KingGeorge
Well, that might be a problem to understanding....
anonymous
  • 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
  • anonymous
Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.
anonymous
  • anonymous
*greater or equal
ParthKohli
  • ParthKohli
That is not what I am able to understand... how?
KingGeorge
  • KingGeorge
Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.
anonymous
  • anonymous
Ah ok
anonymous
  • anonymous
Did you use both constraints when plugging it into wolfram?
KingGeorge
  • KingGeorge
I may have done it a bit wrong though. Let me double check.
anonymous
  • anonymous
Use the Minimize and Maximize routines.
asnaseer
  • 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
  • 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
  • anonymous
I'll give you the link on wolfram alpha in a moment.
asnaseer
  • asnaseer
if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]
anonymous
  • 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
  • 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
  • anonymous
Anyways, the solution using multivariable calculus is perfectly valid and works.
KingGeorge
  • 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
  • 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
  • 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
  • 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
  • asnaseer
I /think/ this works
KingGeorge
  • KingGeorge
Looks good to me.
anonymous
  • anonymous
I dont understand that problem,Parth. Obviously.
asnaseer
  • asnaseer
thx @KingGeorge :)
anonymous
  • anonymous
Hi,asnasser. Greetings,George.
anonymous
  • 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
  • experimentX
how is \[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]
asnaseer
  • 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
  • asnaseer
assuming a and b are both positive
asnaseer
  • asnaseer
hope that helps
asnaseer
  • 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
  • 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
:(
asnaseer
  • asnaseer
where is the flaw in my proof for this?
experimentX
  • 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
  • asnaseer
D'oh! thank for clarifying
asnaseer
  • asnaseer
but does that mean that I have proved the opposite relation to that being asked for?
experimentX
  • experimentX
no ... \[ \sqrt {x-1}+... \ge \sqrt{x-1 + ... } \le \sqrt{x+y+z}\] condition is not strong enough
asnaseer
  • 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
  • anonymous
Why is no one satisfied with treating this as an optimization problem?
asnaseer
  • asnaseer
It's not that I wasn't satisfied - I just wondered if there was a simpler way of arriving at the solution. :)
experimentX
  • experimentX
Lagrange definitely gives solution but I am still looking for more elementary proof.
KingGeorge
  • KingGeorge
There's just something more appealing about the elementary proofs.
anonymous
  • anonymous
I suppose so. At the very least it would be more accessible.

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