Inequality

**gotmejerry** · December 18th 2011, 08:40 AM

(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!

**princeps** · December 18th 2011, 09:20 AM

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!

$1) x+y \geq 2\sqrt {xy}$

$2) x+z \geq 2\sqrt {xz}$

$3) y+z \geq 2\sqrt {yz}$

Now multiply left hand sides and right hand sides of these three equations .

**gotmejerry** · December 18th 2011, 09:23 AM

Thank you!

**princeps** · December 18th 2011, 09:24 AM

you are welcome

**ILikeSerena** · December 18th 2011, 09:29 AM

Just nitpicking...

For x=1, y=-1, z=-1, you get:

$(1-1)(1-1)(-1-1) \ge 8 \cdot 1 \cdot -1 \cdot -1$

$\Rightarrow 0 \ge 8$

which is a contradiction.

**Plato** · December 18th 2011, 09:32 AM

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 $x,~y,~\&~z$ are non-negative.
We have $x^2+y^2\ge 2xy.$ so $x^2z+y^2z\ge 2xyz.$ .

Expand $(x+y)(x+z)(y+z)$ .
Do you see what to do next?

**gotmejerry** · December 18th 2011, 10:03 AM

Sorry, I forgot that x,y,z are postives

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