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Port563
Posts:
673
Registered:
12/15/13


Try and find a geometrical proof for this one, George. (Or, anyone else!)
Posted:
Mar 17, 2014 9:07 PM


At MID <532668f0$0$504$815e3792@news.qwest.net>, George Cornelius very quickly and elegantly found, via wholly geometric means, the minimum value of (x+y)(x+z) where xyz(x+y+z)=1 and x,y,z>0 [all negative would also work].
So, try to find a geometric solution to this one, then... I cannot.
x, y, z > 0 1/3 <= xy + yz + zx <= 3 What are the highest and lowest possible values of each of: (a) xyz (b) x + y + z ?
Failing a geometric solution (which I frankly doubt exists, except perhaps in a very artificial / contrived manner), let's see if an elegant "other" method can be found.
Again, calculus is prohibited.
Intuition may get one to the answer quickly, but a rigorous proof is needed.



