# Show that 0=< x,y,z =< 4 when x+y+z=6, xy+xz+yz=9?

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### 1 Answer

You need to write the sum `x+y` in terms of z such that:

`x + y = 6 - z`

Notice that you may write the next equation such that:

`xy + z*(x + y) = 9`

You should substitute `6 - z` for `x+y` such that:

`xy + z*(6 - z) = 9`

`xy = 9 - z*(6 - z)`

Using the identity (x+y)^2>= 4xy yields:

`(6 - z)^2 gt= 4(9 - z*(6 - z))`

You need to open the brackets such that:

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

`- 12z + z^2 gt=- 24z + 4z^2`

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

`3z^2 - 12z =lt 0`

You need to divide by 3 such that:

`z^2 - 4z =lt 0`

`z^2 - 4z = 0 =gt z(z-4) = 0`

`=gtz = 0`

`=gtz-4 = 0 =gt z = 4`

**Notice that the expression is negative for `z in [0,4], ` hence, reasoning by analogy yields that `x, y in [0,4].` **