1. ## demonstration - help

Demonstrate if x,y,z are > or = than 0.

(x+y)(y+z)(z+x) => 8xyz

"=>" is grater or equal

2. You CAN just beat the tar out of it...

1) Multiply out the LHS.
2) Subtract the 2xyz
3) Divide by xyz
4) Rearrange so that reciprocals are grouped together.
5) Ponder the reciprocal pairs.
6) Prove that any reciprocal pair >= 2

Hint:

f(A) = A + 1/A

What can you tell of that? Maybe a graph would be helpful?

Nothing like a little algebra practrice.

3. Originally Posted by voidstuff
Demonstrate if x,y,z are > or = than 0.

(x+y)(y+z)(z+x) => 8xyz

"=>" is grater or equal
This is a well-known inequality.

We know that

$x,y,z\ge0\implies\begin{cases}x+y&\ge2\sqrt{xy}\\y +z&\ge2\sqrt{yz}\\x+z&\ge2\sqrt{xz}\end{cases}$

Multiply this inequalities to get the desired.

4. thanks a lot... I was trying to solve it with a similar inecuation:

i was using
$x,y,z\ge0\implies\begin{cases}x^2+y^2&\ge2xy\\y^2+ z^2&\ge2yz\\x^2+z^2&\ge2xz\end{cases}$
which was the same thing you proposed but adding $2xy$ on both sides.

gracias amigo chileno xD