Here's the question you clicked on:
ParthKohli
Can anyone help me with this one?
https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash4/389738_558445504198872_734761968_n.jpg Since I am too lazy to write everything down :-)
Don't we also need the requirement that \(x,y,z\ge1\)?
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.
They are anything satisfying \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\)
Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here...-_-
@wmckinely I just know the problem: nothing else ._.
What if you treat it as a minimization/maximization problem.
that's a very good way to look at it.
But the biggest problem is how we proceed.
Never mind, lol. I don't understand that mathematics. :-D
Well, that might be a problem to understanding....
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.
Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.
That is not what I am able to understand... how?
Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.
Did you use both constraints when plugging it into wolfram?
I may have done it a bit wrong though. Let me double check.
Use the Minimize and Maximize routines.
I /think/ you can do this if you use:\[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}\]which I believe can be proved.
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}]
I'll give you the link on wolfram alpha in a moment.
if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]
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
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
Anyways, the solution using multivariable calculus is perfectly valid and works.
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 :/
rearranging slightly we get:\[(x+y+z)^2=(x-1)^2+2x-1+(y-1)^2+2y-1+(z-1)^2+2z-1+4xyz\]\[=(x-1)^2+(y-1)^2+(z-1)^2+2x+2y+2z+4xyz-3\]
then take square roots of both sides to get:\[x+y+z=\sqrt{(x-1)^2+(y-1)^2+(z-1)^2+\text{other terms from above}}\]\[\ge\sqrt{(x-1)^2}+\sqrt{(y-1)^2}+\sqrt{z-1)^2}\]\[\ge(x-1)+(y-1)+(z-1)\]
then square root again to get:\[\sqrt{x+y+z}\ge\sqrt{(x-1)+(y-1)+(z-1)}\]\[\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\]
I dont understand that problem,Parth. Obviously.
Hi,asnasser. Greetings,George.
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.
how is \[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]
\[\begin{align} (\sqrt{a}+\sqrt{b})^2&=a+b+2\sqrt{ab}\\ \therefore a+b&=(\sqrt{a}+\sqrt{b})^2-2\sqrt{ab}\\ &\ge(\sqrt{a}+\sqrt{b})^2\\ \therefore \sqrt{a+b}&\ge\sqrt{a}+\sqrt{b} \end{align}\]
assuming a and b are both positive
it may not be a /strict/ proof but it is one that I used to convince myself that this relation might be valid.
Wait...I'm no longer convinced. Let \(a=9\), \(b=16\). Then \(\sqrt{a+b}=5\) and \(\sqrt{a}+\sqrt{b}=3+4=7>5\).
where is the flaw in my proof for this?
there is a slight error \[ (\sqrt a + \sqrt b)^2 - 2 \sqrt a \sqrt b \leq (\sqrt a + \sqrt b)^2 \] proves the opposite
D'oh! thank for clarifying
but does that mean that I have proved the opposite relation to that being asked for?
no ... \[ \sqrt {x-1}+... \ge \sqrt{x-1 + ... } \le \sqrt{x+y+z}\] condition is not strong enough
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.
Why is no one satisfied with treating this as an optimization problem?
It's not that I wasn't satisfied - I just wondered if there was a simpler way of arriving at the solution. :)
Lagrange definitely gives solution but I am still looking for more elementary proof.
There's just something more appealing about the elementary proofs.
I suppose so. At the very least it would be more accessible.