Algabra (1 Viewer)

[foram]

Prove: x^3 + y^3 + z^3 > 3xyz

Can sombody show me how to do this?

 

[Conspiración]

a) I thought you were "pwning" math?
b) I love how you're doing 3u English and can't spell

I don't know the answers /shrugs

 

[foram]

who needs to spell?

Somebody please help me with this question!

 

[kaz1]

why dont u try completing the cube like (x+y)^3-(3x^2)y-(3y)^2x

 

[tommykins]

Prove: x^3 + y^3 + z^3 > 3xyz

Can sombody show me how to do this?

I'm not sure as I'm pretty cbf at doing it at the moment.

Try making the LHS = (x+y+z)³ by expanding and minusing the results from (x+y+z)³.

ie. a²+b²+c² = (a+b+c)² - 2(ab+ac+bc)

Do it for the cubed, it should cancel out the 3xyz somewhere.

 

[namburger]

Are x,y,z integers?

 

[cwag]

Prove: x^3 + y^3 + z^3 > 3xyz

Can sombody show me how to do this?

what about when x, y, and z or all < 0...this doesn't statement doesn't hold.

 

[foram]

when x and y and z are zero, the whole thing is zero, so it holds true. and teh right hand side can be negative too.

 

[cwag]

when x and y and z are zero, the whole thing is zero, so it holds true. and teh right hand side can be negative too.

< that is a less than sign. negative numbers if you may

 

[Slidey]

Prove: x^3 + y^3 + z^3 > 3xyz

Can sombody show me how to do this?

This is 4unit, not 2unit. Anyway there are 3 ways to solve this: Apply the Cauchy inequality, use induction, or use the arithmet-geometric mean for n=2 and some algebraic manipulation.

1 is beyond the HSC but very easy, 2 is 3unit and above (3unit never does inequalitiy induction, so...), 3 is 4unit level harder inequalities.

Edit: and cwag is correct. Generally one must attach the condition: a, b, c all >= 0.

 

[Slidey]

Still, the question is within the scope of the 3unit syllabus I believe.

Even if inequality induction has never really been tested in 3unit.

 

[martinb]

x^3+y^3+z^3
= (x+y+z)(x^2+y^2+z^2) - (xy+xz+yz)(x+y+z) + (xyz)(3)
= (x+y+z)(x^2+y^2+z^2-[xy+xz+yz]) + 3xyz by factorising the first 2 terms.

We know that (x-y)^2>=0 for all real x and y. It follows that x^2-2xy+y^2>=0, that is x^2+y^2>=2xy.
Similarly, (x-z)^2>=0, that is x^2+z^2>=2xz
and (y-z)^2>=0, that is y^2+z^2>=2yz

Adding the above three inequalities we get:

2x^2+2y^2+2z^2>=2xy+2xz+2yz
x^2+y^2+z^2>=(xy+xz+yz)

Therefore x^2+y^2+z^2-xy-xz-yz>=0

Note that this is the term in the second bracket in the expression for x^3+y^3+z^3:

x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2-[xy+xz+yz]) + 3xyz

Since we know the first term on the RHS must be positive, dropping it will make the LHS larger.

Therefore x^3+y^3+z^3>=3xyz.

 

[lolokay]

that equation only works for x,y,z≥0 right?=/

well, how I solved it was
let x≥0
let y=x+a
let z=x+b
let b≥a≥0

substitute into equation, expand, cancel out, you get
3xa^2 + 3xb^2 + a^3 + b^3 ≥ 3xab
Now x, a and b are ≥ so all terms above are ≥ 0,
For 3xb^2 ≥ 3xab
b ≥ a which is true.
Therefore x^3 + y^3 + z^3 > 3xyz (for x,y,z ≥ 0)