tomas Posted June 11, 2008 Share 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 Link to comment Share on other sites More sharing options...
doG Posted June 11, 2008 Share 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. Link to comment Share on other sites More sharing options...
ellipsis Posted June 20, 2008 Share 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? Link to comment Share on other sites More sharing options...
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