Minimum Value (Posted on 2002-06-19)

    x,y,z > 0
    xy+yz+zx <= 3/4

    P=x+y+z+(1/x)+(1/y)+(1/z)

	Submitted by vohonam
Rating: 3.2500 (8 votes)
Solution:	(Hide)
As it turns out this was correctly solved quite a while ago! See friedlinguini's first comment on this problem (later a different solution is presented by vohonam)

	Subject	Author	Date
	Puzzle Answer	K Sengupta	2022-08-15 07:16:52
	help	Billy Bob	2004-03-28 10:26:31
	Excel Agrees	Charlie	2003-04-01 05:06:02
	Alternative solution (less Maths)	Nigel Nisbet	2003-02-14 08:41:58
	oops	lyeshea	2002-07-02 12:46:36
	possible minimum value	lyeshea	2002-07-02 12:43:56
	re(2): Full solution, I sent to Levik !	levik	2002-06-21 06:03:57
	re: Full solution, I sent to Levik !	TomM	2002-06-20 20:38:06
	Full solution, I sent to Levik !	vohonam	2002-06-20 16:46:11
	HEy !	vohonam	2002-06-20 16:27:24
	Hey !	vohonam	2002-06-20 16:26:34
	No Subject	vohonam	2002-06-20 16:25:23
	Hey !	vohonam	2002-06-20 16:08:10
	This is the better way !	vohonam	2002-06-20 16:03:57
	re(2): Wrong !	Nick Reed	2002-06-20 08:29:03
	re: Wrong !	friedlinguini	2002-06-20 08:25:25
	Wrong !	vohonam	2002-06-20 08:05:49
	Evil!	friedlinguini	2002-06-20 03:58:51