Inequality

great_math · #1 $Old$ 10-07-2008, 11:21 AM

If x,y and z are non-negative reals prove that:

$\frac{\frac{x \sqrt y + y \sqrt z + z \sqrt x}{3} + \frac{y \sqrt x + z \sqrt y + x \sqrt z}{3}}{2} \le \sqrt{\left(\frac{x+y}{2}\right) \left( \frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}$

Greengoblin · #2 $Old$ 10-07-2008, 02:02 PM

Looks like you could reduce it to the AM-GM inequality.

Moo · #3 $Old$ 10-07-2008, 02:13 PM

Quote:

Originally Posted by Greengoblin $View Post$

Looks like you could reduce it to the AM-GM inequality.

AM-GM inequality would give an inequality in the other direction, namely $\geq$

Greengoblin · #4 $Old$ 10-07-2008, 02:23 PM

oh. oops!

I only just glanced at this, sorry

Greengoblin · #5 $Old$ 10-08-2008, 03:50 AM

Quote:

Originally Posted by great_math $View Post$

If x,y and z are non-negative reals prove that:

$\frac{\frac{x \sqrt y + y \sqrt z + z \sqrt x}{3} + \frac{y \sqrt x + z \sqrt y + x \sqrt z}{3}}{2} \le \sqrt{\left(\frac{x+y}{2}\right) \left( \frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}$

I think you jut have to simplyfy both sides down, like so:

$\frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}+y\sqrt{x}+z\sqrt{y}+x\sqrt{z}}{6}\le \sqrt{\frac{(x+y)(y+z)(z+x)}{2^3}}$
$\frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}+y\sqrt{x}+z\sqrt{y}+x\sqrt{z}}{6}\le \sqrt{\frac{(y^2+xy+xz+yz)(z+x)}{8}}$
$\le \sqrt{\frac{zy^2+xy^2+xyz+x^2y+xz^2+x^2z+yz^2+xyz}{8}}$

then try and square both sides, and reduce both numerators as much as possible. My guess is they will turn out the same (but I can't be bothered to do it all), so you will have something like $\frac{u}{36}\le\frac{u}{8}$ where u is an expression in x,y, and z.