# A very hard prove

1. Nov 21, 2007

### inferi

hi,
there is a challenging problem that no one could answer it,so here it is if any one wants to try:
if x,z,y are all positive real numbers prove that:

(1+x^2)(1+y^2)(1+z^2)/(xyz)>=8

so anyone can help please? thank you.

2. Nov 21, 2007

### arildno

Hmm...you could try to start out with:
$$(1-x)^{2}(1-y)^{2}(1-z)^{2}\geq{0}$$
This can be rearranged as:
$$((1+x^{2})-2x)((1+y^{2})-2y)((1+z^{2})-2z)\geq{0}$$
Perhaps this might yield something

3. Nov 21, 2007

### paniurelis

Hello,

I am not a native english speaker, so be warned :-).
This problem is simple really.
For any real number we have
(1-x)^2>=0, so 1+x^2>=2x;
(1-y)^2>=0, so 1+y^2>=2y;
(1-z)^2>=0, so 1+z^2>=2z.

Lets plug these inequalities into the left part of inequality we want to prove:
(1+x^2)(1+y^2)(1+z^2)/xyz >= (2x * 2y * 2z)/xyz=2*2*2=8.
So, we have (1+x^2)(1+y^2)(1+z^2)/(xyz)>=8.
Proved :-)
Hope, this helps.

4. Nov 21, 2007

### arildno

Dear, oh dear, you're right. How simple..

Here's another one:
We may rewrite the the expression as:
$$f(x,y,z)=g(x)*g(y)*g(z), g(x)=x+\frac{1}{x}$$
Thus, the minimum value will occur at:
$$g'(x)=g'(y)=g'(z)=0\to{x}=y=z=1$$

Last edited: Nov 21, 2007
5. Nov 21, 2007

### arildno

Didn't the paniure give that proof already, Kummer?

6. Nov 21, 2007

### Kummer

Yes he did. I should be more careful lest someone accuse me of stealing answers. I deleted my post otherwise it does not look good.

7. Nov 21, 2007

### arildno

Hmm..more likely, it was the not-bothering-reading-all-the-posts syndrome you were the victim of.

I believe you are forgiven.

8. Nov 21, 2007

### therector24

it's simple, just assume that each one is 1 or more so it will work for all of the integers from 1 and above