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@amistre64 @.Sam. ?

Don't we also need the requirement that \(x,y,z\ge1\)?

Yes.

Are they integers?

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.

http://en.wikipedia.org/wiki/Lagrange_multiplier

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....

Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.

*greater or equal

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.

Ah ok

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'll give you the link on wolfram alpha in a moment.

if you multiply both sides of your equation by xyz you get:\[yz+xz+xy=2xyz\]

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.

I /think/ this works

Looks good to me.

I dont understand that problem,Parth.
Obviously.

thx @KingGeorge :)

Hi,asnasser. Greetings,George.

how is
\[\sqrt{a+b}\ge\sqrt{a}+\sqrt{b} \]

assuming a and b are both positive

hope that helps

:(

where is the flaw in my proof for this?

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

Why is no one satisfied with treating this as an optimization problem?

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.