If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
I think it is trivial because they are distinct.
So I wonder just saying "Since they are distinct" is enough to prove it?
Of course there could be several more detailed versions but I just want to know that reasoning is true or not.
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ I don't know why it is trivial. $\endgroup$ – Tunococ May 28 '13 at 8:15
Add a comment
|
$\begingroup$
$\endgroup$
3
HINT: $$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2-3ab\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ $$=(a+b+c)\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2$$
which will be $>=<0 $ according as $a+b+c>=<0$ as for real distinct $a,b,c,$ each of $(a-b)^2,(b-c)^2,(c-a)^2>0$
answered May 28 '13 at 8:18
-
$\begingroup$ Yes, I know that. But then we need a condition $a+b+c \neq 0$, right? So... you mean we have to identify that condition when we want to prove that? $\endgroup$ – noname May 28 '13 at 8:19
-
1$\begingroup$ @noname, the condition supplied is not sufficient $\endgroup$ – lab bhattacharjee May 28 '13 at 8:22
-
$\begingroup$ With $b=-a$, $c=0$ (a simple way to fulfil $a+b+c=0$ with thee different reals), you'll get $-a^3-(-a)^3-0^3+3\cdot 3\cdot (-3)\cdot 0=-a^3+a^3=0$. $\endgroup$ – celtschk May 28 '13 at 8:25
$\begingroup$
$\endgroup$
You statement is wrong, see $(0,-1,1)$.
$\begingroup$
$\endgroup$
$$ (-1)^3 + (-2)^3 + 3^3 = -1 - 8 + 27 = 18 = 3(-1)(-2)(3) $$
$\begingroup$
$\endgroup$
4
Note: I failed to notice that the problem as stated allows $a,b$, and $c$ to be arbitrary real numbers. As others have noted, in this generality the result is false. It is, however, true if $a,b,c\ge 0$. If $-a^3-b^3-c^3+3abc=0$, then
$$\frac13\left(a^3+b^3+c^3\right)=abc=\sqrt[3]{a^3b^3c^3}\;,$$
i.e., the arithmetic mean of $a^3,b^3$, and $c^3$ is the same as their geometric mean. Now use the full form of the AM-GM inequality.
I’ll leave this up for now in case it turns out that the problem was actually supposed to include the hypothesis that $a,b,c\ge 0$.
answered May 28 '13 at 8:19
-
$\begingroup$ The numbers are allowed to be negative, and AM-GM only works for nonnegative numbers. $\endgroup$ – Potato May 28 '13 at 8:21
-
$\begingroup$ @Potato: Ah, I missed the $R$. I wonder if the OP’s source restricted $a,b$, and $c$; it’s hard to see how the problem would arise otherwise. $\endgroup$ – Brian M. Scott May 28 '13 at 8:22
-
$\begingroup$ It must have. The problem is wrong as stated. Counterexamples are given in other answers. $\endgroup$ – Potato May 28 '13 at 8:23
-
$\begingroup$ @Potato: Well, there is the alternative possibility that whoever posed the problem was being sneaky. $\endgroup$ – Brian M. Scott May 28 '13 at 8:24
