Funny Stuff ! (Posted on 2002-06-20)

   a,b,c >0 and a+b+c=1 ;
   P=(1/a)+(2/b)+(3/c);

	Submitted by vohonam
Rating: 3.4286 (7 votes)
Solution:	(Hide)
P*1 = (a+b+c)*(1/a+2/b+3/c); P*1=6+2a/b + 3a/c + b/a + 3b/c + c/a + 2c/b P=6+(2a/b+b/a)+(3a/c+c/a)+(3b/c+2c/b) Use Cauchy's inequality for each : we have : P >= 6+ 2*sqrt(2)+2*sqrt(3)+2*sqrt(6) "=" when a+b+c=1 and b=a*sqrt(2) c=a*sqrt(3) ===>we can solve for a,b,c No maximum value.

	Subject	Author	Date
	re(6): Puzzle Answer	K Sengupta	2022-09-05 14:41:33
	re(5): Puzzle Answer	Steven Lord	2022-09-05 13:10:17
	re(4): Puzzle Answer	K Sengupta	2022-09-05 12:39:30
	re(3): Puzzle Answer	Steven Lord	2022-09-05 11:46:29
	re(2): Puzzle Answer	K Sengupta	2022-09-05 08:33:02
	re: Puzzle Answer	Steven Lord	2022-09-05 05:43:10
	Puzzle Answer	K Sengupta	2022-09-05 03:27:00
	Hard and grusome method	Bon	2004-08-06 04:38:33
	re: Cauchy's inequality !	friedlinguini	2002-06-21 15:17:48
	Cauchy's inequality !	vohonam	2002-06-21 13:58:48
	re: A better way to solve this !	friedlinguini	2002-06-21 12:44:15
	A better way to solve this !	vohonam	2002-06-21 11:58:55
	MINUM = WRONG MAXIMUM = INIFINITY	quddous behrouzi	2002-06-21 10:18:07
	MINIMUM = 27 MAXIMUM = INIFINITY	quddous behrouzi	2002-06-21 10:07:53
	re: Whatever happened to that textbook...?	friedlinguini	2002-06-21 06:00:04
	Whatever happened to that textbook...?	TomM	2002-06-20 22:30:34
	Oh Well	TomM	2002-06-20 20:47:41
	re: solution (maybe)- No	vohonam	2002-06-20 20:29:52
	solution (maybe)	TomM	2002-06-20 20:09:57