Let a,b,c be real numbers. Is a,b,c≥0 the necessary and sufficient condition to show that a3+b3+c3≥3abc?
If not, find the condition that is both sufficient and necessary.
I have just seen Sengupta's excellent solution.
However, it has one small problem.
a
3+b
3+c
3- 3abc = (a+b+c) (a^2+b^2+c^2 –ab-bc-ca) *
It is not true that "(a+b+c)> =0 whenever
lhs of (*) is > = 0"
That is because (a^2+b^2+c^2 –ab-bc-ca) might be 0.
This can only happen if a = b = c.
Whenever a < 0 and a =b = c, it is still true that a
3+b
3+c
3≥3abc, even though (a + b + c ) < = 0.
For example, (a,b,c) = (-2, -2, -2)
SO, THE NECESSARY AND SUFFICIENT CONDITION IS
(a + b + c) >= 0 OR (a = b = c)
The problem would not need this minor modification if it was a strict inequality.
It is true that (a
3+b
3+c
3 > 3abc) if and only if (a + b + c) > 0
Puzzle Solution -- minor correction
