show that

You 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)-2^1/3-2^2/3 = 0

Hence (x-2)³ - (2^1/3)³-(2^2/3)³ = 3 (x-2)2^1/32^2/3 or

(x-2)³-2-4 = 6(x-2)

Simplify this and you should get -6.

or 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)

@prophetsir: pls explain the MUIRHEAD inequality..

smbody explain muirhead inequality///

This 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)

solution to question 2

given, x = 2 + 2^1/3 + 2^2/3

or, x = 2^1/3 (1 + 2^1/3 + 2^2/3)
or, x³ = 2 (2 + 2^1/3 + 2^2/3 - 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....