More fun with Algebra

[Ferdinando]

Prove that for all positive x, y and z

 

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

[the tree]

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.

[uncool]

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...

[chemhelper]

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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[Ragib]

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

[sinisterwolf]

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

[Ragib]

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..

[sinisterwolf]

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?

[Ragib]

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.

[daniel_haxby]

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]

 

QED