tomas Posted June 11, 2008 Posted June 11, 2008 (edited) [math]x[/math] , [math]y[/math],[math]z[/math] non negative reals number and [math]x+y+z=1[/math] . prove that [math]0 \le xy+yz+zx-2xyz \le \frac{7}{27}[/math] Edited June 11, 2008 by tomas
doG Posted June 11, 2008 Posted June 11, 2008 Wow, that looks like a homework question straight out of algebra. Is it? You can solve inequalities like this using linear functions.
ellipsis Posted June 20, 2008 Posted June 20, 2008 Hmm... I think the left half is easier to prove. I think you can just move "-2xyz" over and change it into a good form. But I'm a bit stuck on the right half of the inequality. I was thinking of using ab <= 2(ab)^(1/2)? The condition for that is a=b. Applying this to the inequality means that x=y=z.... This is not a good method at all.... anyone with better ideas?
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