Help?

More general approach (though I've just illustrated the final part graphically rather than describing it formally):

if all x =1 z=1 y=1 xy+xz+yz=3 so 3-xyz=2 9/4=2.25. so 2.25>2!

There are many solutions to this. For instance :

If   x = 1 , y = 2 and z = 0

x + y + z = 3

And

xy + xz + yz - xyz =

2 +  0   +  0  -  0   =   2  ≤  9/4

Switching x and y values and letting z remain 0  would work, as well

I thought zoka123 would be looking for a general proof.    

Thanks Alan,

I did not think it was a simple answer.

I played around with it but I didn't really get anywhere :)

Begin by expanding the expression $\displaystyle(1-x)(1-y)(1-z).$

$\displaystyle (1-x)(1-y)(1-z) = 1-x-y-z+xy+yz+zx-xyz,$

so that

$\displaystyle xy+yz+zx-xyz=x+y+z-1+(1-x)(1-y)(1-z),$

and applying the constraint,

$xy+yz+zx-xyz=2+(1-x)(1-y)(1-z)$

and the  problem becomes one of finding the maximum value of $\displaystyle (1-x)(1-y)(1-z),$ subject to the constraint.

Note that $\displaystyle x=y=z=1$ makes this expression, call it $\displaystyle S,$ equal to zero.

Moving away from these values, all three cannot be greater than or less than 1 since this violates the constraint, making one of them greater than 1 and the other two less than 1 causes $\displaystyle S$ to be negative, while making one of them less than 1 and the other two greater than 1 makes $\displaystyle S$ positive, which is what we want.

Suppose then wlog that $x<1, y>1 \text{ and }z>1.$ 

Substituting for x from the constraint,

$\displaystyle S = (y+z-2)(1-y)(1-z)=(y+z-2)(y-1)(z-1)$.

For $S$ to be as big as possible, we would like y and z to be as big as possible and this will be the case when $y+z=3$, (from the constraint, when $x=0$ ).

Substituting $\displaystyle y=3-z$ into the expression for $\displaystyle S$ produces

$\displaystyle S = (1)(z-2)(1-z)=-(z^{2}-3z+2)=-\left[(z-3/2)^{2}-1/4\right]$,

from which it follows that $\displaystyle S$ has a maximum value of 1/4 occurring when z = 3/2 (and y = 3/2).

The final variable could have been chosen as x or y rather than z,  so 

$\displaystyle xy+yz+zx-xyz\leq2+1/4$

occurring when anyone of x, y or z is equal to zero and the other two equal to 3/2.