# Given relationships x+y+z=6, xy+xz+yz=9, demonstrate 0<=x<=4, 0<=y<=4, 0<=z<=4?

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You need to rewrite the equations provided by the problem, such that:

`{(x + y = 6 - z),(xy = 9 - xz - yz):}`

`{(x + y = 6 - z),(xy = 9 - z(x + y)):}`

Replacing `6 - z` for `x + y` yields:

`{(x + y = 6 - z),(xy = 9 - z(6 - z)):}`

Considering the special inequality `(x + y)^2 >= 4xy` and replacing `6 - z ` for `x + y` andÂ `9 - z(6 - z)` for xy yields:

`(6 - z)^2 >= 4(9 - z(6 - z))`

Expanding the square and performing the calculations, yields:

`36 - 12z + z^2 >= 36 - 24z + 4z^2`

Reducing duplicate terms yields:

`z^2 - 12z >= 4z^2 - 24z => z^2 - 4z^2 + 24z - 12z >= 0`

`-3z^2 + 12z >= 0`

Dividing by -3 yields:

`z^2 - 4z <= 0 => z(z - 4) <= 0`

You should notice that the inequalit holds for values of `z in [0,4]` . Reasoning by analogy yields that `x,y in [0,4]` .

**Hence, testing if `x,y,z in [0,4]` , using the equations provided by the problems, yields that the statement "`x,y,z in [0,4]` " holds.**