# More fun with Algebra

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Prove that for all positive x, y and z

(x+y)^z + (y+z)^x + (z+x)^y > 2

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O.k., it isn't early in the morning this time, so I should be o.k.

For extremely small values of x, y and z: (x+y)z + (y+z)x + (z+x)y is positive.

As x,y or z tend to toward infinity, the whole function also does.

Now how to write that formally, is beyond me.

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Clearly, all 3 have to be less than 1. Then, at least two of them must be less than 1/2, as otherwise we have 1 + (1/2)^y + (1/2)^z where y, z < 1. Let y, z < 1/2. That should start off the problem...

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• 5 weeks later...

Presume that x,y and z are infinitesimally small. Essentially, you will get 0^0 + 0^0 + 0^0. Any quanity, regardless of what it is, raised to the zero power is equal to one by definition. Thus, you will have 1 + 1 + 1 > 2 when x,y,z > 0

This is a basic idea but I'm not certain if it is the actual proof

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Omfg, Fun with ALGEBRA? For a formal proof you need 3 dimensional calculus...at least the proof i can think ok...

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Prove that for all positive x, y and z

(x+y)^z + (y+z)^x + (z+x)^y > 2

Hmmm...I can prove this wrong right now...

3([(0.1)+(0.1)]^2)=0.12<2

reword this and maybe it will work....well work gor a 100% if reworded properly

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I don't think that is correct sinisterwolf, can you give me your values for x y and z seperately? That doesn't look right..

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• 3 weeks later...
I don't think that is correct sinisterwolf, can you give me your values for x y and z seperately? That doesn't look right..

Yeah sorry about that I didnt take a really hard look at that and i thougt that it said "[(x+y)^z]+[(x+y)^z]+[(x+y)^z]" sorry about that I should have taken a better look before i jumped at it.... "make sure brain is running before you engage the mouth" is somthing I hear a lot. But I'll work on this a little bit and come back to try again. Question does x,y and z have to have three seperate values?

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Well it says it has to be proven for all positive values, and I can't say the answers have to be the same, so I would assume they can be separate values.

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$x,y,z = 1$ implies $2+2+2 > 2$

$x,y,z > 1$ implies $f(x,y,z) > 2$

Ok, now let

$x,y,z = 1/N; N \in R^+$ implies $f(x,y,z) = 3(2/N)^{1/N}$

Now, $(2/N)^{1/N} > 2/3$ for $N > 1$, so $f(x,y,z) > 2$ for all $x=y=z$.

Then, whatever $x,y,z$, pick $N$ such that $1/N < min\{x,y,z\}$.

We know $2 < f(1/N,1/N,1/N) < f(x,y,z)$

QED

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