Re: inequality


"Mate" <mmatica@xxxxxxxxxxx> wrote in message news:1162840959.459732.203480@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Dirk Van de moortel wrote:
"d-viper" <jerry144@xxxxx> wrote in message news:4110301.1162825052165.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxxxxx
how to proove such an inequality x^3+y^3+z^3>=x^2*y+y^2*z+z^2*x?

This will only work for x, y, z >= 0.

Using Thomas Mautsch' trick:

Assume that x <= y <= z.
Then put y = x+a and z = x+a+b with a, b >= 0

Then
x^3 + y^3 + z^3 - x^2*y - y^2*z - z^2*x
= ...
= a^3 + b^3 + 2 a^2 b + 2a^2 x + 2 b^2 x + 3 b^2 a + 2 a b x
= 0
since all coefficients are non-negative.

Dirk Vdm

The inequality is not symmetric, so you cannot assume x <= y <= z.

f(x,y,z) = x^3 + y^3 + z^3 - x^2 y - y^2 z - z^2 x
then
f(x,y,z) = f(y,z,x) = f(z,x,y)
and
f(z,y,x) = f(y,x,z) = f(x,z,y)

1) Assume that x <= y <= z.
Then put y = x+a and z = x+a+b with a, b >= 0
Then
x^3 + y^3 + z^3 - x^2*y - y^2*z - z^2*x
= ...
= a^3 + b^3 + 2 a^2 b + 2a^2 x + 2 b^2 x + 3 b^2 a + 2 a b x
= 0
since all coefficients are non-negative.

2) Assume that y <= z <= x.
See 1, since f(x,y,z) = f(y,z,x) = f(z,x,y)

3) Assume that y <= z <= x.
See 1, since f(x,y,z) = f(y,z,x) = f(z,x,y)

4) Assume that z <= y <= x.
Then put y = z+a and x = z+a+b with a, b >= 0
Then
x^3 + y^3 + z^3 - x^2*y - y^2*z - z^2*x
= ...
= a^3 + b^3 + 2 a^2 z + a^2 b + 2 b^2 z + 2 a b^2 + 2 z a b
= 0
since all coefficients are non-negative.

5) Assume that y <= x <= z.
See 4, since f(z,y,x) = f(y,x,z) = f(x,z,y)

6) Assume that x <= z <= y.
See 4, since f(z,y,x) = f(y,x,z) = f(x,z,y)

Dirk Vdm


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