Find all possible real triplet(s) (x,y,z) that satisfy the following system of equations:

     3 = √x + √y + √z,
     3 = x√x + y√y + z√z,
     3 = x2√x + y2√y + z2√z.

(I assume we are talking about realvalued square roots, so x,y,z>=0?)

Using the cauchy schwartz inequality for reals we can obtain:

9=3*3=(x²sqrtx + y²sqrty + z²sqrtz)(sqrtx + sqrty + sqrtz) >=(xsqrtx + ysqrty + zsqrtz)^2=9

Thus we must have equality in cauchy schwartz, which means that there are constants a and b, not both zero,  such that:

ax²sqrtx+bsqrtx=0, ay²sqrty+bsqrty=0, az²sqrtz+bsqrtz=0

If x,y,z is non zero then dividing each inequality by sqrt of x,y,z respectively shows that -b/a=x=y=z->x=y=z=1 (neither of a nor b can be zero, since that would imply both being zero since we assumed zx,y,z non zero). Assuming wlog that x=0 and y,z>0, then y=z by similar argument, but eq. 1 yields y=z=(3/2)^2, while eq. 2 yields y=z=(3/2)^2/3, which is impossible. Also if wlog x,y=0, z>0 we obtain the contradiction that both z=9 and z=9^(1/3), and obviously all of x,y,z cant be zero.

thus only solution is x=y=z=1.

Edited on October 13, 2008, 3:59 pm

Edited on October 13, 2008, 4:03 pm

Edited on October 13, 2008, 4:04 pm

Posted by Jonathan Lindgren on 2008-10-13 15:57:59