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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

Best Answer 

 #5
avatar+30022 
+10

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

 

proof

 Nov 29, 2015
 #2
avatar+284 
+5

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

 Nov 26, 2015
 #3
avatar+111321 
+10

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

 

 

cool cool cool

 Nov 26, 2015
 #4
avatar+109518 
+5

I thought zoka123 would be looking for a general proof.    frown

 Nov 27, 2015
 #5
avatar+30022 
+10
Best Answer

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

 

proof

Alan Nov 29, 2015
 #6
avatar+109518 
0

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
 #7
avatar
+5

 

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

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