Easy Inequality (1 Viewer)

[Harimau]

Its easy, and i cant do it. Hence im an idiot. deduce that x^3 + y^3 + z^3 > 3xyz

And while im here, can i please ask the expansion of (x+y+z)^3? i always forget...

 

[underthesun]

In one of the HSC questions, it was given that:

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^3 - ab - bc - ca)

That probably came from the expansion, so work it out from there

 

[turtle_2468]

The nicest way I think would be to do the following:
1) shift all to LHS. ie you want to prove x^3+y^3+z^3-3xyz>0
2) factorise to wanting to prove (x+y+z)(x^2+y^2+z^2-xy-yz-xz)>0
3) note that x+y+z>0
4) see that (x^2-2xy+y^2)>0, and the same for (x^2-2xz+z^2)>0 and the other one
5) add up these 3 inequalities, then divide by 2 to get the second bracket
6) as both factors >=0, inequality is true.

There are various other ways by calculus etc, but I don't like them. Not really normal inequalities methods..

(x+y+z)^3=x^3+y^3+z^3+2(xy+yz+xz)

 

[ND]

Use AM >= GM.

edit: heh 3 posts at exactly the same time.

 

[Harimau]

Wow thanks guys.