An algebric inequality

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November 10th 2008, 11:44 PM
bkarpuz

An algebric inequality

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.
November 11th 2008, 12:07 AM
NonCommAlg

Quote:

Originally Posted by bkarpuz

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 $\geq$ ?
November 11th 2008, 12:09 AM
bkarpuz

Quote:

Originally Posted by NonCommAlg

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.
November 11th 2008, 12:52 AM
NonCommAlg

Quote:

Originally Posted by bkarpuz

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:

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

imagine how many inequalities you can create using only Chebyshev's inequality + AM-GM inequality! i guess i'm back to high school again! (Mmm)
November 11th 2008, 01:03 AM
bkarpuz

Quote:

Originally Posted by NonCommAlg

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:

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

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

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.
November 11th 2008, 01:32 AM
NonCommAlg

Quote:

Originally Posted by bkarpuz

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 $6, 1, 1.$