Prove inequality

[MM]

Prove that [tex]a^{3}+b^{3}+c^{3} \ge 3abc[/tex]. When does the equality occur? Is this inequality true only for positive numbers?

[adamsmith]

Solution:prove a^3+b^3+c^3>=3abc
we know that a^3+b^3+c^3-3abc=(a+b+c)(a²+b²+c²-ab-ac-bc)
if, (a+b+c)(a²+b²+c²-ab-ac-bc)>0
then a^3+b^3+c^3-3abc>=0
a^3+b^3+c^3>=3abc hence proved

More information on Proving Inequalities http://goo.gl/bukZk8

[Guest]

Proving [tex]a^3+b^3+c^3\geq 3abc[/tex] is equivalent to proving [tex]a^3+b^3+c^3 - 3abc\geq 0[/tex].
As pointed out in the previous post [tex]a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)[/tex]
We can further rearrange this to:
[tex]a^3+b^3+c^3 - 3abc = (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)/2[/tex]

Clearly the right hand bracket is zero only when [tex]a=b=c[/tex] and strictly positive otherwise. So the inequality will hold if either [tex]a+b+c\geq 0[/tex] or [tex]a=b=c[/tex], with equality when [tex]a+b+c=0[/tex] or [tex]a=b=c[/tex].

Hope this helped.

R. Baber