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CaptainBlack · #3 June 25th, 2007, 04:41 AM

Quote:

Originally Posted by red_dog

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.

Q3. For $x,y,z \in \mathbb{R}$ is the inequality tight, if not can you find a tight version.

RonL