1x+y+z=w ='a'
(xy+yz+zx)a=xyz
let (xy+yz+zx)='b'
xyz='c'
ab=c
x,y,z are roots of equation
t3-at2+bt-c=0
t2(t-a)+b(t-a)=0
(t-a)(t2+b)=0
so
x=a,y=-z;(real root)
or
(t-a)(t-i√b)(t+i√b)=0
x=a;y=i√b,z=-i√b
First look this might seem an olympiad problem that may scare some of you...
but try this problem to get an insight on how to solve such problems...
x+y+z = w
1/x + 1/y + 1/z = 1/w
Solve for x, y and z... (Not that you have to find x=1, y=2 , z=3) (Give the answer for real solutions in parametric form)
PS: all you need to do is look at the equation carefully..
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25 Answers
62yes so if y+z=0,
what is xy+yz+zx = x(y+z) + yz
= 0 + y.(-y) = -y2
[1]
62can you complete the y and z part.. there is a slight bit remaining there..
I have given a pink because this much is really good
Infact awesome [1]
162Now there is another method.. (a way to think..)
infact 2 more.. one is prophet sir's approach.. another is mine.. . soumik has factorized it.. but is the factorisation even simpler?
1oh ! I thought this Soumik has answered.. is anything left to be done ??
62Well this one is getting carried for too long.. :(
try this...
Can we show that
xyz=(x+y+z)(xy+yz+zx)??
Now what?
1oops yes i didnt read the ques properly it has asked for all soutions ..
341while that does give infinitely many solutions, it does not tell us we have exhausted all possibilities
1hey guys .. why dont you just put y=-z and x=w ???!! then you fill find out that there are really infinite solutions :P
11I don't see anything of this.....
Only thing I see is \boxed{(x+y)(y+z)(z+x)=0}
Done! [1]
62It seems no one has...
This is something everyone knows.. but doesn't know the name..
Can someone see a polynomial? *(btw when I gave the question I had a different proof in mind)
341i am going to annoy nishant sir and give a hint: Francois Viete
106so, sir we have to find all real solutions in parametric form. ? then my steps till now are correct
62Yes.. I did mean that it has to be found in a parametric form (but din want to give that away..)
I should have used the phrase "Find all real solutions"
106CASE II. All negative solutions
Then in this case too we can replace x,y,z,w with -a,-b,-c,-d where a,b,c,d > 0 and apply AM-HM and reach at the same contradiction
Hence no all negative solutions
So there remain two more cases, 2 negative one positive and 1 negative two positive.
thinking abt them
341@fibonacci - there are infinitely many solutions. Perhaps Nishant sir meant you can express them in some parametric form
106USING AM-HM
(x+y+z)(1/x+1/y+1/z) > 9
=> w.1/w > 9
=> 1 > 9
which is a contradiction
So, there is no positive value of x,y,z for which this is possible