(x+y)*(x+z)*(y+z)>=8xyz How to show this, I suppose somehow with the AM–GM inequality, but don't know how. Thank you!
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Originally Posted by gotmejerry (x+y)*(x+z)*(y+z)>=8xyz How to show this, I suppose somehow with the AM–GM inequality, but don't know how. Thank you! $\displaystyle 1) x+y \geq 2\sqrt {xy}$ $\displaystyle 2) x+z \geq 2\sqrt {xz}$ $\displaystyle 3) y+z \geq 2\sqrt {yz}$ Now multiply left hand sides and right hand sides of these three equations .
Thank you!
you are welcome
Just nitpicking... For x=1, y=-1, z=-1, you get: $\displaystyle (1-1)(1-1)(-1-1) \ge 8 \cdot 1 \cdot -1 \cdot -1 $ $\displaystyle \Rightarrow 0 \ge 8$ which is a contradiction.
Originally Posted by gotmejerry (x+y)*(x+z)*(y+z)>=8xyz How to show this, I suppose somehow with the AM–GM inequality, but don't know how. I assume that $\displaystyle x,~y,~\&~z$ are non-negative. We have $\displaystyle x^2+y^2\ge 2xy.$ so $\displaystyle x^2z+y^2z\ge 2xyz.$. Expand $\displaystyle (x+y)(x+z)(y+z)$. Do you see what to do next?
Sorry, I forgot that x,y,z are postives
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