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 = x**^{2}√x + y^{2}√y + z^{2}√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=(x**^{2}sqrtx + y^{2}sqrty + z^{2}sqrtz)(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:**

**
****ax**^{2}sqrtx+bsqrtx=0, **ay**^{2}sqrty+bsqrty=0, **az**^{2}sqrtz+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**