1. a4 + b4 + c4 > 3abc (a + b + c)
2. if x = 2 + 21/3 + 22/3, then find x3 - 6x3 + 6x
Please please please......... solve these....
341You seem to be pretty careless in posting your questions.
1) is obviously wrong for a=b=c. The right inequality is:
a4+b4+c4 ≥ abc(a+b+c). Some of you may like to take a shot at proving this.
2) You must be looking for x3-6x2+6x
We have (x-2)-21/3-22/3 = 0
Hence (x-2)3 - (21/3)3-(22/3)3 = 3 (x-2)21/322/3 or
(x-2)3-2-4 = 6(x-2)
Simplify this and you should get -6.
36I m sorry for that......
I did it last night at 3 and so......
The correct question is....
a4 + b4 + c4 > abc(a + b + c) and i want a proof of this.
This is a question from Junior mathematics olympiad 2009
The answer of the second one is 2... but don't solve that if u don't want to...
But please solve the inequality one...
341or Rearrangement Inequality or Muirhead.
Proof by AM-GM:
a4+a4+b4+c4≥4a2bc
a4+b4+b4+c4≥4ab2c
a4+b4+c4+c4≥4abc2
Adding the three, we get
4(a4+b4+c4)≥4abc(a+b+c) or
a4+b4+c4≥abc(a+b+c)
21from tchebychef's inequality n(a1*b1 + a2*b2 +....+ an*bn)≥(a1+a2+...+an)(b1+b2+...+bn)
NOW we can write 3(a4+b4+c4)≥(a3+b3+c3)(a+b+c)
or,a4+b4+c4≥(a3+b3+c3)(a+b+c)/3
further from AM GM (a3+b3+c3)/3≥abc
so (a4+b4+c4)≥ abc(a+b+c)
21from rearrangement inequality finally we appear at the tchebychefs step...is their any shorter solution???
341This is an accessible reference:
http://www.artofproblemsolving.com/Wiki/index.php/Muirhead's_Inequality
First understand the term majorizes which is a pretty easy concept.
So the sequence (4,0,0) majorizes (2,1,1)
Hence
\sum x^4 = \sum_{sym} x^4y^0z^0 \ge \sum_{sym} x^2y^1z^1 = xyz(x+y+z)
36solution to question 2
given, x = 2 + 21/3 + 22/3
or, x = 21/3 (1 + 21/3 + 22/3)
or, x3 = 2 (2 + 21/3 + 22/3 - 1)3
or, x3 = 2 (x - 1)3
or, x3 = 2 (x3 - 1 - 3x2 + 3x)
or, x3 = 2x3 - 2 - 6x2 + 6x
or, x3 - 6x2 + 6x = 2 Ans....
1I am impressed Shubhodip[1]....Got to learn a lot from everybody out there.
