# Help with the following inequality

1. Aug 14, 2008

### evagelos

.........................[(x+y)(y+z)(z+x) +xy+xz+yz]/xyz >=
.........................(3xy+3xz+3yz+x^2+y^2+z^2)/[(x+y)(y+z)(z+x)]

2. Aug 14, 2008

### snipez90

Re: inequalities

That doesn't look like a very appealing inequality. Is there a condition for this inequality? I guess you could try dumbassing it by assuming x >= y >= z since the inequality is symmetric in x,y,z and then cross multiplying but I haven't figured out a nice solution yet.

3. Aug 14, 2008

### evagelos

Re: inequalities

thanks for the quick respond the only condition is that x>0,y>0,z>0

4. Aug 14, 2008

### morphism

Re: inequalities

What have you tried?

5. Aug 15, 2008

### cellotim

Re: inequalities

I would suggest, if x,y,z are positive, that you just get a common denominator and multiply the numerators out into expanded polynomials, then match up the terms.

6. Aug 15, 2008

### snipez90

Re: inequalities

Yeah just as I had guessed, the inequality is basically just multiplying out, moving everything to one side, and relying on the fact that the sum of positive real numbers is greater than 0.

That is really a nasty inequality. It's symmetric but that doesn't make anything better. It's not homogeneous, so we can't come up with any additional constraints. It doesn't rely on any clever manipulations or other well-known inequalities (unless you count a simple albeit important fact). It's somewhat instructive but taken in perspective it's no where near as instructive as the other inequality you posed.

7. Aug 15, 2008

### evagelos

Re: inequalities

so what do we do now?

8. Aug 15, 2008

### evagelos

Re: inequalities

Give me a solid proof please

9. Aug 16, 2008

### cellotim

Re: inequalities

The change of variables:

$$s = x + y + z, t = xyz, w = 1/x + 1/y + 1/z$$

might help. Note that $$(x + y)(y + z)(x + z) = swt - t$$ and $$xy + xz + yz = tw$$. The inequality becomes: $$sw + w - 1 \geq (s^2 + 2tw)/(stw - t)$$, if my calculations aren't off. Also note that $$sw \geq 1$$.

10. Aug 16, 2008

### evagelos

Re: inequalities

cellotim you mixed up worst now

11. Aug 17, 2008

### cellotim

Re: inequalities

Does this mean that I made a mistake or that I've confused you more?

If you want to know how I would solve this problem for a class, I would sit down at Maple or Mathematica, cross-multiply the denominators, and have the computer symbolically expand each side (in x,y,z), then I would make sure that all the coefficients on the LHS are bigger than all the coefficients on the RHS. It would be a huge mess, but it would get it done. I don't know if you have access to any maths software, but that is how I would do it. (Of course you can do it all or partially by hand if you need to show that work. It wouldn't take more than 30-1hr.)