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Algebra
22 Mar 2008 10:25:55 IST
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[-2,1] then prove that x2+y2+z2 
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sandeep ramesh
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22 Mar 2008 10:31:59 IST
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maybe this will work
Its easily proved that only one number can be less than -1
So let that be z WLOG
So make it 2(x^2 + y^2 + xy) <= 6
This is easily proved since |x| and |y| <=1 :)
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22 Mar 2008 10:52:46 IST
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sandeep what did u do after u got one of the has to be less than -1???
can u explain a bit
here is my proof
let x,y,z be the roots of a cubic
P(b)=b^{3}+ub-v
u=xy+yz+zx
all it's roots lie within [-2,1]
thus the roots of
P'(b)=0 alpha,eta also lie within [-2,1]
so alpha,=+sqrt{-u/3}<=1
thus u+3 >=0 or 6>=-2u=0^{2}-2u=x^{2}+y^{2}+x^{2}
can u explain a bit
here is my proof
let x,y,z be the roots of a cubic
P(b)=b^{3}+ub-v
u=xy+yz+zx
all it's roots lie within [-2,1]
thus the roots of
P'(b)=0 alpha,eta also lie within [-2,1]
so alpha,=+sqrt{-u/3}<=1
thus u+3 >=0 or 6>=-2u=0^{2}-2u=x^{2}+y^{2}+x^{2}
22 Mar 2008 10:57:04 IST
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my soln was:
Case 1: WLOG |z| >= 1
Implies there can be no other number with |no| >1
So z = -x -y
implies
the expr is (x+y)^2 + x^2 + y^2 = 2(x^2 + y^2 + xy)
But |x| and |y| <=1
implies the expr <= 6
Case 2: | No number| > =1
Same proof :)
ie
the expr is (x+y)^2 + x^2 + y^2 = 2(x^2 + y^2 + xy)
But |x| and |y| <=1
implies the expr <= 6
[-2,1] 
22 Mar 2008 11:13:06 IST
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x^2+y^2+z^2<=6
x^2+y^2+z^2=-2(xy+yz+xz)
by rearrangement
the laest value
y^2+2xz
this should be as negative as possible
x<y<z
x is positive , z is negative
min y^2=0
value
-2 x=1,z=-1
let us add any positive quantity to y and sub it from z as x cant increase
q^2+2(-1-q)
(q-1)^2-3
whose min is q=1
-3
x^2+y^2+z^2<=-2*-3=6
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