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suppose this equality holds:

x, y, z are real

15(x + y + z) = 12(xy + yz + xz) = 10(x^2 + y^2 + z^2 )

and at least one variables isn't zero.

I need to proof that x + y + z = 4 and find the smallest posible closed interval [a,b] in which are all possible solutions for x, y, z.

I have tried several approaches, but all of them went to some non-sense.

Thanks.

P.S. For some strange reason I can't add the greeting to the beginning of this post, so sorry.

balast
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1 Answers1

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Hint

Using the identity

$$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$$

and the given equalities you have

$$(x+y+z)^2=\frac{15}{10}(x+y+z)+\frac{15}{6}(x+y+z).$$

Could you continue from this point?

mfl
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  • Cool, that helped with x + y + z = 4 problem, but what about the interval? Now I know these: x+y+z = 4, xy + yz + xz = 5, x^2 + y^2 + z^2 = 6. – balast Oct 19 '14 at 19:43
  • $x+y+z=4$ is a plane and $x^2+y^2+z^2=6$ is a sphere. The intersection is a circumference. Could you get the center and the radius? Maybe this link http://www.ambrsoft.com/TrigoCalc/Spher/SpherePlaneIntersection_.htm could be useful. – mfl Oct 19 '14 at 19:48
  • I can solve it using your link or WolframAlpha, but isn't there a simplier way? – balast Oct 19 '14 at 20:32
  • With your particular data you could try to find a center of the form $(a,a,a)$ to get $a=0.$ But, at this moment, I don't know a way to get the answer more quickly. – mfl Oct 19 '14 at 20:43
  • I asked the same question on reddit and some reditor is helping me out as well, so maybe you could look there… – balast Oct 19 '14 at 21:04
  • Are you allowed to use Lagrange multipliers? – mfl Oct 19 '14 at 21:36
  • I have no idea what that is… – balast Oct 19 '14 at 21:39