an inequality Watch

cxs
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#1
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given that x,y,z are are distinct and non-zero, and they satisfy
x+1/y=y+1/z=z+1/x
show that x^2y^2z^2=1 and that the value of x+1/y is either 1 or -1
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DFranklin
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A good starting point is to write x-y in terms of y and z.
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cxs
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i use it to figure out i,but for ii,i suppose it is m,by write m^3and 3m, i got that m could be1,-1,2,-2———noidea how to leave out the2and-2
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cxs
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sorry i find that i still have no idea how to answer the part2(my method is to write m=x+1/y, while i can get the relationship beween m^3 and 3m, but donno how to leave out 2 and -2, and i think this method is not good, for if it changes to another question similiarly, perhaps i will not have the luck to find relation between m^3 and 3m.
(Original post by DFranklin)
A good starting point is to write x-y in terms of y and z.
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cxs
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how stupid i am
just solve it when it is all 2or -2will reveal that they are equal!
well originally i think there's one way of using an inequality to prove that they are equal when at that time…
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DFranklin
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(Original post by cxs)
how stupid i am
just solve it when it is all 2or -2will reveal that they are equal!
well originally i think there's one way of using an inequality to prove that they are equal when at that time…
Have to admit, I got stuck at the same point as you; I did mean to post to at least let you know I was stuck too, but other things kept taking priority and the thread fell off my notification list.

Glad you got something out in the end (I'd be curious to see what your eventual solution was, but don't bother if it's a lot of work to type up).
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cxs
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here's my final solution (still using my original methods with only one point refined.)
Spoiler:
Show
x+1/y=y+1/z => x-y=1/z-1/y=(y-z)/yz
so yz=(y-z)/(x-y) it is beautiful, fot it is symmetric. the same we have xy=..., xz=...
so x2y2z2=1. and xyz=1or-1
let m=x+1/y=y+1/z=z+1/x, then m3=xyz+1/xyz+(x+1/y)+(y+1/z)+(z+1/x)=2+3m when xyz=1 and -2+3m when xyz=-1
solve this we get m=1,-1,2,-2
to leave out 2,-2. suppose m=2, then x+1/y=y+1/z=z+1/x=2
(then what stuck me first is that although we can add them to get (x+1/x)+(y+1/y)+(z+1/z)=6, and these three are no smaller than 2 if they are all positive----but idk how to prove they are all positive,until---)
note that when m=2, there's another condition that xyz=1 (as m=2 is derived from xyz=-1)
they are all positive or two of them are negative. without loss of generality, suppose x<0, y<0, then x+1/y can not be 2, so contradiction,now we can add them, and they are all positive, each (x+1/x) is 2, so x=y=z=1, and they are not equal, so leave out 2(the same we can leave out -2)

still I think there must be other methods.
(Original post by DFranklin)
Have to admit, I got stuck at the same point as you; I did mean to post to at least let you know I was stuck too, but other things kept taking priority and the thread fell off my notification list.

Glad you got something out in the end (I'd be curious to see what your eventual solution was, but don't bother if it's a lot of work to type up).
P.S. I didn't reply soon cos your reply time is almost bed time for me (in my country)~~
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