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

#Algebra #Inequalities

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Comments

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 - 6 years, 1 month 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?

Figel Ilham - 6 years, 1 month 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:

$a+b > 0, (a-b) ^2 > 0 \Rightarrow ( a + b)(a-b) ^2 > 0$ .
$\sum (a+b)(a-b)^2 > 0 \Rightarrow 2( \sum a^3) > \sum ab(a+b)$ 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 - 6 years, 1 month ago

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

Figel Ilham - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$