Thread: Problem 28
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Old June 25th, 2007, 01:47 AM
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If x,y,z\in\mathbf{R} then
1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow
\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow
\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}. So the proposition is true.
If x,y,z\in\mathbf{C} then let x=i,y=-i,z=1.
Then \displaystyle xy+xz+yz=1>\frac{1}{2}. So the proposition is false.