# Inequality

• Dec 18th 2011, 08:40 AM
gotmejerry
Inequality
(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!

• Dec 18th 2011, 09:20 AM
princeps
Re: Inequality
Quote:

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 .
• Dec 18th 2011, 09:23 AM
gotmejerry
Re: Inequality
Thank you!
• Dec 18th 2011, 09:24 AM
princeps
Re: Inequality
you are welcome
• Dec 18th 2011, 09:29 AM
ILikeSerena
Re: Inequality
Just nitpicking...

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

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

$\displaystyle \Rightarrow 0 \ge 8$

• Dec 18th 2011, 09:32 AM
Plato
Re: Inequality
Quote:

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?
• Dec 18th 2011, 10:03 AM
gotmejerry
Re: Inequality
Sorry, I forgot that x,y,z are postives:D