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.
If c = 0, then the equation becomes
a
3+b
3≥0 ?
In this case, it is necessary and sufficient that (a + b) > 0
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If b = c = 1, then the equation becomes
a
3+2
≥ 3a
A little graphing shows that it is necessary and sufficient that a > -2
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If b = 1, c = 2, then the equation becomes
a
3+9 ≥ 6a
A little graphing shows that it is necessary and sufficient that a > -3
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If b = -1, c = -1, then the equation becomes
a
3 - 2 ≥ 3a
A little graphing shows that it is necessary and sufficient that a > 2
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A suspicion is that is necessary and sufficient that
(a+b+c) >= 0.
This seems possible. Let a = -(b+c)
Then a
3+b
3+c
3 =
-(b+c)
3+b
3+c
3=
-3b*c*b-3b*c*c =
3(-b-c)bc = 3abc
More thoughts; Solution?
