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More fun with Algebra

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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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  • 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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  • 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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How about this:


[math]x,y,z = 1 [/math] implies [math]2+2+2 > 2[/math]

[math]x,y,z > 1 [/math] implies [math]f(x,y,z) > 2[/math]


Ok, now let


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


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


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


We know [math]2 < f(1/N,1/N,1/N) < f(x,y,z)[/math]



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