How can I demonstrate that (a+b+c)(a³+b³+c³-3abc) is greater than 0 for all a,b,c real numbers

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ChickenMeister2

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How can I demonstrate that (a+b+c)(a³+b³+c³-3abc) is greater than 0 for all a,b,c real numbers

How can I demonstrate that?
 
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Subhotosh Khan said:
What happens when a = b = c = 1?

Please show us what you have tried and exactly where you are stuck.

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In general what do get for the value of the expression below:

(a+b+c)(a³+b³+c³-3abc)​


when

a = b = c
 
Sorry guys, it seems like it can be greater than or equal to 0, the exercise had a mistake. But I still can't find a way to demonstrate it for every (a, b,c)
 
This is a help forum where we help students solve their problems. Where do you need help? Can you post what you have done so far? If you had read the posting guidelines you would have received help by now.

Hint: Multiply out (a+b+c)^3 or possibly (a+b+c)^4
 
\(\displaystyle (a+b+c)(a^3+b^3+c^3-3abc)=(a+b+c)^2(a^2+b^2+c^2-ab-bc-ca)\\\)
\(\displaystyle =(a+b+c)^2\times\frac{1}{2}(2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca)\\ \)
\(\displaystyle =(a+b+c)^2\times\frac{1}{2}\{(a^2 + b^2 – 2ab) + (b^2 + c^2 – 2bc) + (c^2 + a^2 – 2ca)\}\\\)
\(\displaystyle =(a+b+c)^2\times\frac{1}{2}\{(a - b)^2 + (b - c)^2 + (c – a)^2\}\ge 0\\\)
The expression is the product and sum of squares. Therefore, it is greater or equal to 0 for all values of [imath]a\,,b\,,\,c[/imath].
 
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