Ferdinando Posted December 4, 2006 Share Posted December 4, 2006 Prove that for all positive x, y and z (x+y)^z + (y+z)^x + (z+x)^y > 2 Link to comment Share on other sites More sharing options...

the tree Posted December 4, 2006 Share Posted December 4, 2006 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. Link to comment Share on other sites More sharing options...

uncool Posted December 10, 2006 Share Posted December 10, 2006 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... Link to comment Share on other sites More sharing options...

chemhelper Posted January 9, 2007 Share Posted January 9, 2007 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 ----- Have homework questions in math, physics or chemistry? Who Likes Homework -- http://www.wholikeshomework.com Link to comment Share on other sites More sharing options...

Ragib Posted January 11, 2007 Share Posted January 11, 2007 Omfg, Fun with ALGEBRA? For a formal proof you need 3 dimensional calculus...at least the proof i can think ok... Link to comment Share on other sites More sharing options...

sinisterwolf Posted January 13, 2007 Share Posted January 13, 2007 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 Link to comment Share on other sites More sharing options...

Ragib Posted January 13, 2007 Share Posted January 13, 2007 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.. Link to comment Share on other sites More sharing options...

sinisterwolf Posted January 30, 2007 Share Posted January 30, 2007 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? Link to comment Share on other sites More sharing options...

Ragib Posted January 30, 2007 Share Posted January 30, 2007 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. Link to comment Share on other sites More sharing options...

daniel_haxby Posted January 30, 2007 Share Posted January 30, 2007 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 Link to comment Share on other sites More sharing options...

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