Let a,b,c be real numbers. Is a,b,c≥0 the necessary and sufficient condition to show that a^{3}+b^{3}+c^{3}≥3abc?
If not, find the condition that is both sufficient and necessary.

Since each of a, b and c are real, it follows that:

(a-b)^2 > = 0, (b-c)^2 > = 0 and (c-a)^2 > = 0, so that:

(a^2+b^2+c^2 –ab-bc-ca)

= 1/2*{(a-b)^2 + (b-c)^2 + (c-a)^2)

> = 0

Now, we know that:

a^3+b^3+c^3 – 3*a*b*c = (a+b+c) (a^2+b^2+c^2 –ab-bc-ca) ……(*)

Now, a^2+b^2+c^2 –ab-bc-ca> = 0. Thus, if a+b+c >= 0, then it follows that the lhs of (*) is > = 0

If lhs of (*) > 0, then (a^2+b^2+c^2 –ab-bc-ca) cannot be 0, since that would force a^3+b^3+c^3 – 3*a*b*c = 0, a contradiction.

Hence, (a^2+b^2+c^2 –ab-bc-ca)> 0 in terms of (i), so that a+b+c > 0

If lhs of (*) is 0, then:

Either, a+b+c = 0

Or, a^2+b^2+c^2 –ab-bc-ca= 0

-> 1/2*{(a-b)^2 + (b-c)^2 + (c-a)^2) = 0.

-> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0

Since each of a, b and c are real, it follows that each of (a-b)^2, (b-c)^2 and (c-a)^2 must be equal to 0

Accordingly, a=b=c

Consequently, the condition that is both sufficient and necessary would be:

a+b+c >=0, OR a=b=c

*Edited on ***February 28, 2008, 5:19 am**