ou_litu
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It is about inequality, which is shown as follows.
Thanks for help.
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ou_litu
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RDKGames
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(Original post by ou_litu)
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This is the same as proving

9(a^3 + b^3 + c^3) - (a+b+c)^3 \geq 0

So expand the LHS. Try to show this is the case.
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ou_litu
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(Original post by RDKGames)
This is the same as proving

9(a^3 + b^3 + c^3) - (a+b+c)^3 \geq 0

So expand the LHS. Try to show this is the case.
I tried to solve it in your way, but after expansion, no progress at all.
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Gregorius
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(Original post by ou_litu)
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Jensen's inequality will give you this very quickly. If you're not allowed to assume Jensen, then follow the method of proof!
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RDKGames
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(Original post by ou_litu)
I tried to solve it in your way, but after expansion, no progress at all.
Alright, it was just an idea. Though looking at the question from a source, I have to ask, is there a previous part to your question where you prove the inequality

4(a^3+b^3) \geq (a+b)^3 ??

Because it seems to be useful.
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ou_litu
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(Original post by RDKGames)
Alright, it was just an idea. Though looking at the question from a source, I have to ask, is there a previous part to your question where you prove the inequality

4(a^3+b^3) \geq (a+b)^3 ??

Because it seems to be useful.
Sorry, still can't see your pount.
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username2844924
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Another way:

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3(a^3+a^3+b^3+b^3+c^3+c^3) >= 3(a^2b+b^2c+c^2a+b^2a+c^2b+a^2c) by the rearrangement inequality

a^3+b^3+c^3 >= 3abc by AM-GM so 2(a^3+b^3+c^3) >= 3abc

Add these together to get:

8(a^3+b^3+c^3) >= 3(a^2b+b^2c+c^2a+b^2a+c^2b+a^2c+abc)

so 9(a^3+b^3+c^3) >= 3(a^2b+b^2c+c^2a+b^2a+c^2b+a^2c+abc) + a^3+b^3+c^3 = (a+b+c)^3.

Last edited by username2844924; 2 years ago
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RDKGames
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(Original post by ou_litu)
Sorry, still can't see your pount.
So is that a yes to my question? If so, notice that from this inequality we have that

4(a^3+b^3) \geq (a+b)^3
4(b^3+c^3) \geq (b+c)^3
4(c^3+a^3) \geq (c+a)^3

Add them up:

8(a^3+b^3+c^3) \geq 2(a^3+b^3+c^3) + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3c^2a + 3ca^2

\implies 9(a^3 + b^3 + c^3) \geq 2(a^3+b^3+c^3) + 3(a^2b +ab^2 +b^2c + bc^2 + c^2a + ca^2) + (a^3+b^3+c^3)

If you can move from there.
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ou_litu
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(Original post by RDKGames)
So is that a yes to my question? If so, notice that from this inequality we have that

4(a^3+b^3) \geq (a+b)^3
4(b^3+c^3) \geq (b+c)^3
4(c^3+a^3) \geq (c+a)^3

Add them up:

8(a^3+b^3+c^3) \geq 2(a^3+b^3+c^3) + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3c^2a + 3ca^2

\implies 9(a^3 + b^3 + c^3) \geq 2(a^3+b^3+c^3) + 3(a^2b +ab^2 +b^2c + bc^2 + c^2a + ca^2) + (a^3+b^3+c^3)

If you can move from there.
Can't imagine it can be solved in such an easy way.
Thanks a lot.
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ou_litu
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(Original post by J843126028)
Another way:

Spoiler:
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3(a^3+a^3+b^3+b^3+c^3+c^3) >= 3(a^2b+b^2c+c^2a+b^2a+c^2b+a^2c) by the rearrangement inequality

a^3+b^3+c^3 >= 3abc by AM-GM so 2(a^3+b^3+c^3) >= 3abc

Add these together to get:

8(a^3+b^3+c^3) >= 3(a^2b+b^2c+c^2a+b^2a+c^2b+a^2c+abc)

so 9(a^3+b^3+c^3) >= 3(a^2b+b^2c+c^2a+b^2a+c^2b+a^2c+abc) + a^3+b^3+c^3 = (a+b+c)^3.


Thanks.
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