Strange inequality (1 Viewer)

[J-Wang]

given and , prove

Anyone know how to do this?

 

[Carrotsticks]

2 ways I can see, here's the start. Try what you can.

- Harmonic Mean.

- Cross multiply so you have the cyclic sum on top and the product on the bottom.

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=given, x@plus;y@plus;z\geq 3\sqrt[3]{xyz}\\ let x=1/a,y=1/b,z=1/c\\ \therefore \frac{1}{a}@plus;\frac{1}{b}@plus;\frac{1}{c}\geq \frac{3}{\sqrt[3]{abc}}\\ now,~(x@plus;y@plus;z)/3\geq \sqrt[3]{xyz}\\ 1/3\geq \sqrt[3]{xyz}\\ 3/\sqrt[3]{xyz}\geq 9\\ put~together~to~yield~result." target="_blank"><img src="http://latex.codecogs.com/gif.latex?given, x+y+z\geq 3\sqrt[3]{xyz}\\ let x=1/a,y=1/b,z=1/c\\ \therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3}{\sqrt[3]{abc}}\\ now,~(x+y+z)/3\geq \sqrt[3]{xyz}\\ 1/3\geq \sqrt[3]{xyz}\\ 3/\sqrt[3]{xyz}\geq 9\\ put~together~to~yield~result." title="given, x+y+z\geq 3\sqrt[3]{xyz}\\ let x=1/a,y=1/b,z=1/c\\ \therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3}{\sqrt[3]{abc}}\\ now,~(x+y+z)/3\geq \sqrt[3]{xyz}\\ 1/3\geq \sqrt[3]{xyz}\\ 3/\sqrt[3]{xyz}\geq 9\\ put~together~to~yield~result." /></a>
oh yeh, and this was for all a,b,c>0 of course