harder 3u question (1 Viewer)

[cutemouse]

Suppose that (1+x)(1+y)(1+z)=8. Prove that xyz<= 1

Thanks

 

[Timothy.Siu]

Suppose that (1+x)(1+y)(1+z)=8. Prove that xyz<= 1

Thanks

x,y,z are integers right?

 

[cutemouse]

Yep... x>0, y>0, and z>0 -- Sorry should've said it before.

 

[lolokay]

expand, you get 1 + x + y + z + xy + yz + zx + xyz = 8
using the fact (for numbers >= 0) a + b >= 2sqrt(ab) for each pair of eg. x and yz we get:
6sqrt(xyz) + xyz =< 7
let sqrt(xyz) = a, then we have
a² + 6a - 7 <= 0
(a+7)(a-1) <= 0
as a is positive, a =< 1 for this to be true
so sqrt(xyz) =< 1
xyz =< 1

 

[Trebla]

 

[lolokay]

^ ah yes, i missed the 1+xyz pair

 

[cutemouse]

Does the fact that the previous question (or part) said that x+y+z >= 3(xyz)^(1/3) help?

 

[omniscience]

Suppose that (1+x)(1+y)(1+z)=8. Prove that xyz<= 1

Thanks

not hard enough