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# Help?

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if x+y+z=3 and x≥0 , y≥0 , z≥0

Prove that  xy + xz + yz - xyz ≤ 9/4

Nov 25, 2015

#5
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More general approach (though I've just illustrated the final part graphically rather than describing it formally):

Nov 29, 2015

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

Nov 26, 2015
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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

Nov 26, 2015
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I thought zoka123 would be looking for a general proof.

Nov 27, 2015
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More general approach (though I've just illustrated the final part graphically rather than describing it formally):

Alan Nov 29, 2015
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Thanks Alan,

I did not think it was a simple answer.

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

Nov 30, 2015
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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.

Nov 30, 2015