Prove that cuberoot(xyz).(1/x + 1/y + 1/z) >= 3 where x,y,z > 0 and are elements of real?

Let u = 1/x, v = 1/y and w = 1/z. Rearrange to get

     (u + v + w) / 3 ≥ ³√(u v w )

which follows from the AM-GM inequality.

Sorry, don't have a lot of time to respond, but as is the case with many inequalities, you may have to prove it in pieces, by taking cases. Prove it first for the easier cases, where 0 < x, y, z < 1 and where x, y, z > 1 (it's fairly obvious for these cases). Then take two more cases, where one variable is between 0 and 1 and the other two are > 1, and the fourth case, where two variables are between 0 and 1 and the third is > 1. Not as obvious how the proof will work in that case but that's how I would tackle it. Finally, show that equality is attained if x, y, z all equal 1.

Hope this helps at least a bit.

x²/xz + y²/yx + z²/zy >= (x + y + z)²/(xz+yx+zy) (Cauchy-Schwarz in Engel type) yet (x+y+z)²/(xz+yx+zy) >= (x+y+z) iff (x+y+z)² >= (x+y+z)(xz + yx + zy) >= (3sqrt(xyz))² (cauchy-schwarz) = 9xyz x+y+z>=3cbrt(xyz)>=3sqrt(xyz) hence shown edit returned: particularly we are able to coach the equivalent subject of three>=x+y+z (when you consider that x+y+z>=3cbrt(xyz)): x/z+yx+z/y >= 3 >= x+y+z edit: sorry I merely found out I misstated "equivalent"... yet i've got considered an analogous subject with the subject 3>=x+y+z which, of direction, implies your given subject (yet weaker). EDIT returned: oooops i'm so sorry i assume i replaced into snoozing while i replaced into doing this... the coolest judgment in my first answer is incorrect... :( i will attempt to come back up with yet another one...