K kanali New member Joined Jun 12, 2010 Messages 18 Jun 12, 2010 #1 Prove the following inequality: \(\displaystyle xyz\geq x+y+z-2\) for \(\displaystyle x\geq 1,y\geq 1,z\geq 1\). i can only think of squaring the equation but would that help??
Prove the following inequality: \(\displaystyle xyz\geq x+y+z-2\) for \(\displaystyle x\geq 1,y\geq 1,z\geq 1\). i can only think of squaring the equation but would that help??
D Deleted member 4993 Guest Jun 15, 2010 #2 kanali said: Prove the following inequality: \(\displaystyle xyz\geq x+y+z-2\) for \(\displaystyle x\geq 1,y\geq 1,z\geq 1\). i can only think of squaring the equation but would that help?? Click to expand... I have not totally thought it through - but I think following identity might help: \(\displaystyle a^3 \ \ + \ \ b^3 \ \ + \ \ c^3 \ \ - \ \ 3abc \ \ = \ \ (a \ \ + \ \ b \ \ + \ \ c)\cdot (a^2 \ \ + \ \ b^2 \ \ + \ \ c^2 \ \ - \ \ ab \ \ - \ \ bc \ \ - \ \ ca)\)
kanali said: Prove the following inequality: \(\displaystyle xyz\geq x+y+z-2\) for \(\displaystyle x\geq 1,y\geq 1,z\geq 1\). i can only think of squaring the equation but would that help?? Click to expand... I have not totally thought it through - but I think following identity might help: \(\displaystyle a^3 \ \ + \ \ b^3 \ \ + \ \ c^3 \ \ - \ \ 3abc \ \ = \ \ (a \ \ + \ \ b \ \ + \ \ c)\cdot (a^2 \ \ + \ \ b^2 \ \ + \ \ c^2 \ \ - \ \ ab \ \ - \ \ bc \ \ - \ \ ca)\)
