An algebric inequality

bkarpuz · #1 $Old$ 11-11-2008, 01:44 AM

For $x,y,z>0$ , prove
$x^{5}+y^{5}+z^{5}\leq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2}}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.$

thanks.

NonCommAlg · #2 $Old$ 11-11-2008, 02:07 AM

Quote:

Originally Posted by bkarpuz $View Post$

For $x,y,z>0$ , prove
$x^{5}+y^{5}+z^{5}\geq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2}}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.$

thanks.

the inequality in this form is not true for all positive reals x, y, z, because it fails if you put z = 1, x = y, and then let x go to infinity! are you sure it's not $\leq$ instead of $\geq$ ?

bkarpuz · #3 $Old$ 11-11-2008, 02:09 AM

Quote:

Originally Posted by NonCommAlg $View Post$

the inequality in this form is not true for all positive reals x, y, z, because it fails if you put z = 1, x = y, and then let x go to infinity! are you sure it's not $\leq$ instead of $\geq$ ?

I correct it now.

NonCommAlg · #4 $Old$ 11-11-2008, 02:52 AM

Quote:

Originally Posted by bkarpuz $View Post$

For $x,y,z>0$ , prove
$x^{5}+y^{5}+z^{5}\leq x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2}}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}}.$

thanks.

because of symmetry, we may assume that $x \leq y \leq z.$ then obviously: $\sqrt{\frac{x^{2}}{yz}} \leq \sqrt{\frac{y^{2}}{xz}} \leq \sqrt{\frac{z^{2}}{xy}}.$ hence by Chebyshev's inequality we'll have:

$x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2}}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}} \geq \frac{1}{3}(x^5 + y^5 + z^5) \left(\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\sqrt{\frac{z^{2}}{xy}} \right).$ but by AM-GM we have: $\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\sqrt{\frac{z^{2}}{xy}} \geq 3. \ \ \ \ \ \Box$

imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again!

bkarpuz · #5 $Old$ 11-11-2008, 03:03 AM

Quote:

Originally Posted by NonCommAlg $View Post$

because of symmetry, we may assume that $x \leq y \leq z.$ then obviously: $\sqrt{\frac{x^{2}}{yz}} \leq \sqrt{\frac{y^{2}}{xz}} \leq \sqrt{\frac{z^{2}}{xy}}.$ hence by Chebyshev's inequality we'll have:

$x^{5}\sqrt{\frac{x^{2}}{yz}}+y^{5}\sqrt{\frac{y^{2}}{xz}}+z^{5}\sqrt{\frac{z^{2}}{xy}} \geq \frac{1}{3}(x^5 + y^5 + z^5) \left(\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\sqrt{\frac{z^{2}}{xy}} \right).$ but by AM-GM we have: $\sqrt{\frac{x^{2}}{yz}}+\sqrt{\frac{y^{2}}{xz}}+\sqrt{\frac{z^{2}}{xy}} \geq 3. \ \ \ \ \ \Box$

imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again!

I was thinking of a different solution.
I know that this inequality can be shown by special inequalities, the first one dropped in my mind was Muirhead's inequality (again because of the symmetry), which you may find in Muirhead's inequality - Wikipedia, the free encyclopedia.
If there is any other solutions, I would like to see.

Thanks NonCommAlg.

NonCommAlg · #6 $Old$ 11-11-2008, 03:32 AM

Quote:

Originally Posted by bkarpuz $View Post$

I was thinking of a different solution.
I know that this inequality can be shown by special inequalities, the first one dropped in my mind was Muirhead's inequality (again because of the symmetry), which you may find in Muirhead's inequality - Wikipedia, the free encyclopedia.
If there is any other solutions, I would like to see.

Thanks NonCommAlg.

when i was a high school kid, we were trained to avoid using Muirhead's inequality as much as we can! i never knew why? i guess i'm still under the infulence! lol

anyway, yes, Muirhead's inequality will also solve the problem because if you multiply both sides of the inequality by $xyz$ you'll get: $\sum_{sym}x^6yz \leq \sum_{sym}x^7 \sqrt{y}\sqrt{z},$

which is true because $7, \frac{1}{2}, \frac{1}{2}$ majorizes $6, 1, 1.$