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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
2 years, 10 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?

Staff - 2 years, 10 months ago

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?

- 2 years, 10 months ago

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.

Staff - 2 years, 10 months ago

So, when we write the proof, we should do it in forwards instead of backwards. Thanks a lot, Master :)

- 2 years, 10 months ago