Inequality proving using arithmetical and geometrical mean.

[remix00]

Hello!
I have struggled to prove this inequation. I tried different methods, but i cant even get close to result.
It would be really helpful if someone could give me some hint how to start this right. The thing that should be used here is arithmetical and geometrical mean, but i cant find how to use it here correctly.


1/(x+xy) + 1/(y+yz) + 1/(z+zx) ≥ 3/(1+xyz)


Thanks!

 

[Deleted member 4993]

Hello!
I have struggled to prove this inequation. I tried different methods, but i cant even get close to result.
It would be really helpful if someone could give me some hint how to start this right. The thing that should be used here is arithmetical and geometrical mean, but i cant find how to use it here correctly.


1/(x+xy) + 1/(y+yz) + 1/(z+zx) ≥ 3/(1+xyz)


Thanks!

So let us start at the beginning:

What is the arithmetic mean of (x,y,z)?

What is the geometric mean of (x,y,z)?

 

[remix00]

Arithmetic - (x+y+z)/3
Geometric - [FONT=&quot]^(1/3)xyz

And the inequality would be [/FONT](x+y+z)/3 ≥ [FONT=&quot]^(1/3)xyz = [/FONT][FONT=&quot] [/FONT]x+y+z ≥ 3*[FONT=&quot]^(1/3)xyz[/FONT][FONT=&quot]
But i cant find how to insert it into this inequality to prove it.[/FONT]