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Another inequality (2)

I want to ask also whether this proof is valid.

Given \(a,b,c\) positive real and different, prove that \[2(a^3+b^3+c^3)>bc(b+c) + ca(c+a) +ab(a+b)\]

I proved it using this inequality \[(a+b)(a-b)^2 +(b+c)(b-c)^2 +(c+a)(c-a)^2 >0\]

Since \[(a+b)(a-b)^2>0\] and etc.

Is that valid? Thank you

Note by Figel Ilham
1 year, 8 months ago

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What makes you think that it is valid or invalid? Do you have any argument for either case? Do you have any concerns about either possibility? Calvin Lin Staff · 1 year, 8 months ago

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@Calvin Lin I do have a concern about this arguments \[(a+b)(a-b)^2 \geq0\] since \(a+b\geq 0\) and \((a-b)^2 \geq 0\) I did it by backwards and proof it using the arguments I have. Or is there any other way how to proof that problem? Figel Ilham · 1 year, 8 months ago

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@Figel Ilham What you have shown is that both of these terms are non-negative, and so when we multiply them they are still non-negative. This proof is valid.

A better way of writing up inequality solutions, is to do it in reverse from how you solved it. Namely:

  1. \( a+b > 0, (a-b) ^2 > 0 \Rightarrow ( a + b)(a-b) ^2 > 0 \).
  2. \( \sum (a+b)(a-b)^2 > 0 \Rightarrow 2( \sum a^3) > \sum ab(a+b) \) by expanding.

This way, all that we need is the forward implications, instead of the backwards ones. It now becomes much clearer how to arrive at the solution. Calvin Lin Staff · 1 year, 8 months ago

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@Calvin Lin So, when we write the proof, we should do it in forwards instead of backwards. Thanks a lot, Master :) Figel Ilham · 1 year, 8 months ago

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